Is it possible to perform a computation to know if an object is at rest and its location given a force versus...












1












$begingroup$


The problem is as follows:




The force graphed in Figure 1. is applied to a $textrm{2.0-kg}$
block that was sliding to the right (the $textrm{+-direction}$) over
a frictionless surface with speed $5.0,frac{m}{s}$ at $x=0$ (a) Is
the block ever at rest? If so,where? (b) Find a position (other than
$x=0$) when the block is again moving to the right at
$5.0,frac{m}{s}$.




Sketch of the problem



The figure alluded in the problem is shown above.



Now the thing with this problem is I really don't know where to begin or if there is something missing in my maths.



Hence my question is how do I know that the block is at rest?.



My first guess is that this happens when the net force on the block is $0$ or that it is not carrying (unsure if this term is right) kinetic energy?. However can this happen in a frictionless surface?. My other question arises from the fact, how do I understand the negative force in the diagram?. Does it mean that is a force that is being applied to the left side of the block?.



So far the only thing I could come up with wad to use Newton's second law:



$f=ma$



Since what it is being asked is the position, I thought to use:



$v^{2}_f=v^{2}_{0}+2aDelta x$



For $v^{2}_f=0$ and $v_{0}=5,frac{m}{s}$,



$0=left(5right)^{2}_{0}+2aDelta x$



But this would mean to use a, from where? From the diagram?.



If so, at $-40 N$ :



$a=frac{-40}{2}=-20,frac{m}{s^2}$



Then:



$0=left(5right)^{2}+2left(-20right)Delta x$



Hence:



$Delta x=frac{25}{40}=0.625$



Which would translate as a position $x=0$



$x_{1}-x_{0}=0.625-0=0.625,m$



So this would be at the right from the starting point at $x=0$ and although this seems consistent with what it is alluded in the diagram (the "domain" is in the positive numbers) it doesn't check with the answers in my book.



According to it, the answer for part (a) is $1,textrm{m}$ and for part (b) is $2.917,textrm{m}$. But again, none of this seems close to what I obtained. What could I be missing in my calculation?.



Anyways. Am I understanding this mathematically or physically wrong?. Can somebody help me to clear these doubts?.










share|cite|improve this question











$endgroup$












  • $begingroup$
    "the net force on the block is 0 or that it is not carrying (unsure if this term is right) kinetic energy". You know that these are two entirely different phenomenon, right? Net force = 0 does not imply rest.
    $endgroup$
    – harshit54
    Jan 3 at 18:37












  • $begingroup$
    @harshit54 I'm so sorry. I'm still new to this field. Mind to enlighten me with what you refer to as "different phenomenon"?. My doubt is, there must be kinetic energy since there is a speed $v_0$ but net force must be zero when it comes at rest?. Am I mixing things up?. Maybe you can help me to clear out these questions.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:41










  • $begingroup$
    No need to be sorry. Rest is implied when the velocity of a particle is zero. Velocity can be found out by taking the integral of the acceleration function with respect to time.
    $endgroup$
    – harshit54
    Jan 3 at 19:54










  • $begingroup$
    In your case you just need to calculate the area under the curve from $t= 0$ to some $t=t$ and set it equal to zero. All this is done assuming you start at from rest.
    $endgroup$
    – harshit54
    Jan 3 at 19:58












  • $begingroup$
    @harshit54 Thanks! But then the question arises to, what function to use for integration? as I believe there are three intervals in the graph. Just should I try them all?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:00
















1












$begingroup$


The problem is as follows:




The force graphed in Figure 1. is applied to a $textrm{2.0-kg}$
block that was sliding to the right (the $textrm{+-direction}$) over
a frictionless surface with speed $5.0,frac{m}{s}$ at $x=0$ (a) Is
the block ever at rest? If so,where? (b) Find a position (other than
$x=0$) when the block is again moving to the right at
$5.0,frac{m}{s}$.




Sketch of the problem



The figure alluded in the problem is shown above.



Now the thing with this problem is I really don't know where to begin or if there is something missing in my maths.



Hence my question is how do I know that the block is at rest?.



My first guess is that this happens when the net force on the block is $0$ or that it is not carrying (unsure if this term is right) kinetic energy?. However can this happen in a frictionless surface?. My other question arises from the fact, how do I understand the negative force in the diagram?. Does it mean that is a force that is being applied to the left side of the block?.



So far the only thing I could come up with wad to use Newton's second law:



$f=ma$



Since what it is being asked is the position, I thought to use:



$v^{2}_f=v^{2}_{0}+2aDelta x$



For $v^{2}_f=0$ and $v_{0}=5,frac{m}{s}$,



$0=left(5right)^{2}_{0}+2aDelta x$



But this would mean to use a, from where? From the diagram?.



If so, at $-40 N$ :



$a=frac{-40}{2}=-20,frac{m}{s^2}$



Then:



$0=left(5right)^{2}+2left(-20right)Delta x$



Hence:



$Delta x=frac{25}{40}=0.625$



Which would translate as a position $x=0$



$x_{1}-x_{0}=0.625-0=0.625,m$



So this would be at the right from the starting point at $x=0$ and although this seems consistent with what it is alluded in the diagram (the "domain" is in the positive numbers) it doesn't check with the answers in my book.



According to it, the answer for part (a) is $1,textrm{m}$ and for part (b) is $2.917,textrm{m}$. But again, none of this seems close to what I obtained. What could I be missing in my calculation?.



Anyways. Am I understanding this mathematically or physically wrong?. Can somebody help me to clear these doubts?.










share|cite|improve this question











$endgroup$












  • $begingroup$
    "the net force on the block is 0 or that it is not carrying (unsure if this term is right) kinetic energy". You know that these are two entirely different phenomenon, right? Net force = 0 does not imply rest.
    $endgroup$
    – harshit54
    Jan 3 at 18:37












  • $begingroup$
    @harshit54 I'm so sorry. I'm still new to this field. Mind to enlighten me with what you refer to as "different phenomenon"?. My doubt is, there must be kinetic energy since there is a speed $v_0$ but net force must be zero when it comes at rest?. Am I mixing things up?. Maybe you can help me to clear out these questions.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:41










  • $begingroup$
    No need to be sorry. Rest is implied when the velocity of a particle is zero. Velocity can be found out by taking the integral of the acceleration function with respect to time.
    $endgroup$
    – harshit54
    Jan 3 at 19:54










  • $begingroup$
    In your case you just need to calculate the area under the curve from $t= 0$ to some $t=t$ and set it equal to zero. All this is done assuming you start at from rest.
    $endgroup$
    – harshit54
    Jan 3 at 19:58












  • $begingroup$
    @harshit54 Thanks! But then the question arises to, what function to use for integration? as I believe there are three intervals in the graph. Just should I try them all?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:00














1












1








1





$begingroup$


The problem is as follows:




The force graphed in Figure 1. is applied to a $textrm{2.0-kg}$
block that was sliding to the right (the $textrm{+-direction}$) over
a frictionless surface with speed $5.0,frac{m}{s}$ at $x=0$ (a) Is
the block ever at rest? If so,where? (b) Find a position (other than
$x=0$) when the block is again moving to the right at
$5.0,frac{m}{s}$.




Sketch of the problem



The figure alluded in the problem is shown above.



Now the thing with this problem is I really don't know where to begin or if there is something missing in my maths.



Hence my question is how do I know that the block is at rest?.



My first guess is that this happens when the net force on the block is $0$ or that it is not carrying (unsure if this term is right) kinetic energy?. However can this happen in a frictionless surface?. My other question arises from the fact, how do I understand the negative force in the diagram?. Does it mean that is a force that is being applied to the left side of the block?.



So far the only thing I could come up with wad to use Newton's second law:



$f=ma$



Since what it is being asked is the position, I thought to use:



$v^{2}_f=v^{2}_{0}+2aDelta x$



For $v^{2}_f=0$ and $v_{0}=5,frac{m}{s}$,



$0=left(5right)^{2}_{0}+2aDelta x$



But this would mean to use a, from where? From the diagram?.



If so, at $-40 N$ :



$a=frac{-40}{2}=-20,frac{m}{s^2}$



Then:



$0=left(5right)^{2}+2left(-20right)Delta x$



Hence:



$Delta x=frac{25}{40}=0.625$



Which would translate as a position $x=0$



$x_{1}-x_{0}=0.625-0=0.625,m$



So this would be at the right from the starting point at $x=0$ and although this seems consistent with what it is alluded in the diagram (the "domain" is in the positive numbers) it doesn't check with the answers in my book.



According to it, the answer for part (a) is $1,textrm{m}$ and for part (b) is $2.917,textrm{m}$. But again, none of this seems close to what I obtained. What could I be missing in my calculation?.



Anyways. Am I understanding this mathematically or physically wrong?. Can somebody help me to clear these doubts?.










share|cite|improve this question











$endgroup$




The problem is as follows:




The force graphed in Figure 1. is applied to a $textrm{2.0-kg}$
block that was sliding to the right (the $textrm{+-direction}$) over
a frictionless surface with speed $5.0,frac{m}{s}$ at $x=0$ (a) Is
the block ever at rest? If so,where? (b) Find a position (other than
$x=0$) when the block is again moving to the right at
$5.0,frac{m}{s}$.




Sketch of the problem



The figure alluded in the problem is shown above.



Now the thing with this problem is I really don't know where to begin or if there is something missing in my maths.



Hence my question is how do I know that the block is at rest?.



My first guess is that this happens when the net force on the block is $0$ or that it is not carrying (unsure if this term is right) kinetic energy?. However can this happen in a frictionless surface?. My other question arises from the fact, how do I understand the negative force in the diagram?. Does it mean that is a force that is being applied to the left side of the block?.



So far the only thing I could come up with wad to use Newton's second law:



$f=ma$



Since what it is being asked is the position, I thought to use:



$v^{2}_f=v^{2}_{0}+2aDelta x$



For $v^{2}_f=0$ and $v_{0}=5,frac{m}{s}$,



$0=left(5right)^{2}_{0}+2aDelta x$



But this would mean to use a, from where? From the diagram?.



If so, at $-40 N$ :



$a=frac{-40}{2}=-20,frac{m}{s^2}$



Then:



$0=left(5right)^{2}+2left(-20right)Delta x$



Hence:



$Delta x=frac{25}{40}=0.625$



Which would translate as a position $x=0$



$x_{1}-x_{0}=0.625-0=0.625,m$



So this would be at the right from the starting point at $x=0$ and although this seems consistent with what it is alluded in the diagram (the "domain" is in the positive numbers) it doesn't check with the answers in my book.



According to it, the answer for part (a) is $1,textrm{m}$ and for part (b) is $2.917,textrm{m}$. But again, none of this seems close to what I obtained. What could I be missing in my calculation?.



Anyways. Am I understanding this mathematically or physically wrong?. Can somebody help me to clear these doubts?.







algebra-precalculus physics






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 19:48







Chris Steinbeck Bell

















asked Jan 3 at 18:31









Chris Steinbeck BellChris Steinbeck Bell

855315




855315












  • $begingroup$
    "the net force on the block is 0 or that it is not carrying (unsure if this term is right) kinetic energy". You know that these are two entirely different phenomenon, right? Net force = 0 does not imply rest.
    $endgroup$
    – harshit54
    Jan 3 at 18:37












  • $begingroup$
    @harshit54 I'm so sorry. I'm still new to this field. Mind to enlighten me with what you refer to as "different phenomenon"?. My doubt is, there must be kinetic energy since there is a speed $v_0$ but net force must be zero when it comes at rest?. Am I mixing things up?. Maybe you can help me to clear out these questions.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:41










  • $begingroup$
    No need to be sorry. Rest is implied when the velocity of a particle is zero. Velocity can be found out by taking the integral of the acceleration function with respect to time.
    $endgroup$
    – harshit54
    Jan 3 at 19:54










  • $begingroup$
    In your case you just need to calculate the area under the curve from $t= 0$ to some $t=t$ and set it equal to zero. All this is done assuming you start at from rest.
    $endgroup$
    – harshit54
    Jan 3 at 19:58












  • $begingroup$
    @harshit54 Thanks! But then the question arises to, what function to use for integration? as I believe there are three intervals in the graph. Just should I try them all?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:00


















  • $begingroup$
    "the net force on the block is 0 or that it is not carrying (unsure if this term is right) kinetic energy". You know that these are two entirely different phenomenon, right? Net force = 0 does not imply rest.
    $endgroup$
    – harshit54
    Jan 3 at 18:37












  • $begingroup$
    @harshit54 I'm so sorry. I'm still new to this field. Mind to enlighten me with what you refer to as "different phenomenon"?. My doubt is, there must be kinetic energy since there is a speed $v_0$ but net force must be zero when it comes at rest?. Am I mixing things up?. Maybe you can help me to clear out these questions.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:41










  • $begingroup$
    No need to be sorry. Rest is implied when the velocity of a particle is zero. Velocity can be found out by taking the integral of the acceleration function with respect to time.
    $endgroup$
    – harshit54
    Jan 3 at 19:54










  • $begingroup$
    In your case you just need to calculate the area under the curve from $t= 0$ to some $t=t$ and set it equal to zero. All this is done assuming you start at from rest.
    $endgroup$
    – harshit54
    Jan 3 at 19:58












  • $begingroup$
    @harshit54 Thanks! But then the question arises to, what function to use for integration? as I believe there are three intervals in the graph. Just should I try them all?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:00
















$begingroup$
"the net force on the block is 0 or that it is not carrying (unsure if this term is right) kinetic energy". You know that these are two entirely different phenomenon, right? Net force = 0 does not imply rest.
$endgroup$
– harshit54
Jan 3 at 18:37






$begingroup$
"the net force on the block is 0 or that it is not carrying (unsure if this term is right) kinetic energy". You know that these are two entirely different phenomenon, right? Net force = 0 does not imply rest.
$endgroup$
– harshit54
Jan 3 at 18:37














$begingroup$
@harshit54 I'm so sorry. I'm still new to this field. Mind to enlighten me with what you refer to as "different phenomenon"?. My doubt is, there must be kinetic energy since there is a speed $v_0$ but net force must be zero when it comes at rest?. Am I mixing things up?. Maybe you can help me to clear out these questions.
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 19:41




$begingroup$
@harshit54 I'm so sorry. I'm still new to this field. Mind to enlighten me with what you refer to as "different phenomenon"?. My doubt is, there must be kinetic energy since there is a speed $v_0$ but net force must be zero when it comes at rest?. Am I mixing things up?. Maybe you can help me to clear out these questions.
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 19:41












$begingroup$
No need to be sorry. Rest is implied when the velocity of a particle is zero. Velocity can be found out by taking the integral of the acceleration function with respect to time.
$endgroup$
– harshit54
Jan 3 at 19:54




$begingroup$
No need to be sorry. Rest is implied when the velocity of a particle is zero. Velocity can be found out by taking the integral of the acceleration function with respect to time.
$endgroup$
– harshit54
Jan 3 at 19:54












$begingroup$
In your case you just need to calculate the area under the curve from $t= 0$ to some $t=t$ and set it equal to zero. All this is done assuming you start at from rest.
$endgroup$
– harshit54
Jan 3 at 19:58






$begingroup$
In your case you just need to calculate the area under the curve from $t= 0$ to some $t=t$ and set it equal to zero. All this is done assuming you start at from rest.
$endgroup$
– harshit54
Jan 3 at 19:58














$begingroup$
@harshit54 Thanks! But then the question arises to, what function to use for integration? as I believe there are three intervals in the graph. Just should I try them all?.
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 20:00




$begingroup$
@harshit54 Thanks! But then the question arises to, what function to use for integration? as I believe there are three intervals in the graph. Just should I try them all?.
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 20:00










2 Answers
2






active

oldest

votes


















1












$begingroup$


$v^{2}_f=v^{2}_{0}+2aDelta x$




That applies when you have constant acceleration. The graph shows that force, and therefore acceleration, is changing. So you can't use this formula.



You have that the horizontal axis (width) is distance, and the vertical (height) is force. Height times width is area, and force times distance is work. So area in this diagram represents work (keep in mind that area below the x-axis is negative work). In general, when you have a diagram where the horizontal has units $a$, and the vertical has units $b$, then area represents something with units $ab$.



You're given that at $x=0$, the velocity is +5. From that, and the mass, you can figure out the initial kinetic energy. For the object to be at rest, the new kinetic energy must be 0, which means that work so far must cancel out the initial kinetic energy. So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy.




how do I understand the negative force in the diagram?




A negative force means a force in the "negative" direction. What direction is considered "negative" is somewhat arbitrary, and can be chosen according to what's convenient for the problem. If it's not stated, it's generally assumed that right is positive and left is negative. And indeed, the problem says "the right (the +-direction)", so that tells you that right is the positive direction, and left is the negative direction.






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$endgroup$













  • $begingroup$
    @Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:53










  • $begingroup$
    @ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
    $endgroup$
    – Acccumulation
    Jan 3 at 20:01










  • $begingroup$
    @Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:10










  • $begingroup$
    @Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:38










  • $begingroup$
    @ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
    $endgroup$
    – amd
    Jan 3 at 21:13



















1












$begingroup$

Hint.



You have



$$
m ddot x = f(x) = begin{cases}
5 (6 x-8) & x le 2 \
20 & 2lt x le 3 \
10 (2 x-4) & x gt 3
end{cases}
$$



then



$$
mdot x ddot x = f(x)dot xto frac m2 frac{d}{dt}dot x ^2 = frac{d}{dt}F(x)
$$



then integrating



$$
frac 12 m (v^2-v_0^2) = F(x) = begin{cases}
15 x^2-40 x & xleq 2 \
20 x-60 & 2<xleq 3 \
20 left(frac{x^2}{2}-2 xright)+30 & 3 < x
end{cases}
$$



Attached the plot for $f(x)$ in blue, the plot for $F(x)$ in red and the plot for $v(x)$ in black showing the points in which $v = v_0$ which occurs at $x = 0$ and $x = 3$



enter image description hereenter image description hereenter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:57










  • $begingroup$
    I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 21:04










  • $begingroup$
    @ChrisSteinbeckBell I hope now helps more.
    $endgroup$
    – Cesareo
    Jan 3 at 21:06










  • $begingroup$
    @ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
    $endgroup$
    – Cesareo
    Jan 3 at 21:08






  • 1




    $begingroup$
    @ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
    $endgroup$
    – Andrei
    Jan 4 at 17:02












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2 Answers
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active

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votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$


$v^{2}_f=v^{2}_{0}+2aDelta x$




That applies when you have constant acceleration. The graph shows that force, and therefore acceleration, is changing. So you can't use this formula.



You have that the horizontal axis (width) is distance, and the vertical (height) is force. Height times width is area, and force times distance is work. So area in this diagram represents work (keep in mind that area below the x-axis is negative work). In general, when you have a diagram where the horizontal has units $a$, and the vertical has units $b$, then area represents something with units $ab$.



You're given that at $x=0$, the velocity is +5. From that, and the mass, you can figure out the initial kinetic energy. For the object to be at rest, the new kinetic energy must be 0, which means that work so far must cancel out the initial kinetic energy. So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy.




how do I understand the negative force in the diagram?




A negative force means a force in the "negative" direction. What direction is considered "negative" is somewhat arbitrary, and can be chosen according to what's convenient for the problem. If it's not stated, it's generally assumed that right is positive and left is negative. And indeed, the problem says "the right (the +-direction)", so that tells you that right is the positive direction, and left is the negative direction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:53










  • $begingroup$
    @ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
    $endgroup$
    – Acccumulation
    Jan 3 at 20:01










  • $begingroup$
    @Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:10










  • $begingroup$
    @Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:38










  • $begingroup$
    @ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
    $endgroup$
    – amd
    Jan 3 at 21:13
















1












$begingroup$


$v^{2}_f=v^{2}_{0}+2aDelta x$




That applies when you have constant acceleration. The graph shows that force, and therefore acceleration, is changing. So you can't use this formula.



You have that the horizontal axis (width) is distance, and the vertical (height) is force. Height times width is area, and force times distance is work. So area in this diagram represents work (keep in mind that area below the x-axis is negative work). In general, when you have a diagram where the horizontal has units $a$, and the vertical has units $b$, then area represents something with units $ab$.



You're given that at $x=0$, the velocity is +5. From that, and the mass, you can figure out the initial kinetic energy. For the object to be at rest, the new kinetic energy must be 0, which means that work so far must cancel out the initial kinetic energy. So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy.




how do I understand the negative force in the diagram?




A negative force means a force in the "negative" direction. What direction is considered "negative" is somewhat arbitrary, and can be chosen according to what's convenient for the problem. If it's not stated, it's generally assumed that right is positive and left is negative. And indeed, the problem says "the right (the +-direction)", so that tells you that right is the positive direction, and left is the negative direction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:53










  • $begingroup$
    @ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
    $endgroup$
    – Acccumulation
    Jan 3 at 20:01










  • $begingroup$
    @Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:10










  • $begingroup$
    @Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:38










  • $begingroup$
    @ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
    $endgroup$
    – amd
    Jan 3 at 21:13














1












1








1





$begingroup$


$v^{2}_f=v^{2}_{0}+2aDelta x$




That applies when you have constant acceleration. The graph shows that force, and therefore acceleration, is changing. So you can't use this formula.



You have that the horizontal axis (width) is distance, and the vertical (height) is force. Height times width is area, and force times distance is work. So area in this diagram represents work (keep in mind that area below the x-axis is negative work). In general, when you have a diagram where the horizontal has units $a$, and the vertical has units $b$, then area represents something with units $ab$.



You're given that at $x=0$, the velocity is +5. From that, and the mass, you can figure out the initial kinetic energy. For the object to be at rest, the new kinetic energy must be 0, which means that work so far must cancel out the initial kinetic energy. So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy.




how do I understand the negative force in the diagram?




A negative force means a force in the "negative" direction. What direction is considered "negative" is somewhat arbitrary, and can be chosen according to what's convenient for the problem. If it's not stated, it's generally assumed that right is positive and left is negative. And indeed, the problem says "the right (the +-direction)", so that tells you that right is the positive direction, and left is the negative direction.






share|cite|improve this answer









$endgroup$




$v^{2}_f=v^{2}_{0}+2aDelta x$




That applies when you have constant acceleration. The graph shows that force, and therefore acceleration, is changing. So you can't use this formula.



You have that the horizontal axis (width) is distance, and the vertical (height) is force. Height times width is area, and force times distance is work. So area in this diagram represents work (keep in mind that area below the x-axis is negative work). In general, when you have a diagram where the horizontal has units $a$, and the vertical has units $b$, then area represents something with units $ab$.



You're given that at $x=0$, the velocity is +5. From that, and the mass, you can figure out the initial kinetic energy. For the object to be at rest, the new kinetic energy must be 0, which means that work so far must cancel out the initial kinetic energy. So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy.




how do I understand the negative force in the diagram?




A negative force means a force in the "negative" direction. What direction is considered "negative" is somewhat arbitrary, and can be chosen according to what's convenient for the problem. If it's not stated, it's generally assumed that right is positive and left is negative. And indeed, the problem says "the right (the +-direction)", so that tells you that right is the positive direction, and left is the negative direction.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 3 at 19:47









AcccumulationAcccumulation

7,2052619




7,2052619












  • $begingroup$
    @Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:53










  • $begingroup$
    @ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
    $endgroup$
    – Acccumulation
    Jan 3 at 20:01










  • $begingroup$
    @Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:10










  • $begingroup$
    @Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:38










  • $begingroup$
    @ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
    $endgroup$
    – amd
    Jan 3 at 21:13


















  • $begingroup$
    @Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:53










  • $begingroup$
    @ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
    $endgroup$
    – Acccumulation
    Jan 3 at 20:01










  • $begingroup$
    @Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:10










  • $begingroup$
    @Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 20:38










  • $begingroup$
    @ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
    $endgroup$
    – amd
    Jan 3 at 21:13
















$begingroup$
@Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 19:53




$begingroup$
@Accumulation So you need to find a point on the graph such that area to the left of that point is the negative of the initial kinetic energy. So all that I have to do is to relate both works, a work which cancels the other?. Can you do an example?. Sorry but I am still a bit stuck with this one. I mean if kinetic energy performs work, then other kinetic energy will null the effect of the first (work) and this is understood as the block to be at rest?.
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 19:53












$begingroup$
@ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
$endgroup$
– Acccumulation
Jan 3 at 20:01




$begingroup$
@ChrisSteinbeckBell Work is change in energy. If the initial kinetic energy is $E_0$, then you need to change the energy from $E_0$ to $0$, which is a change of $-E_0$, so you need the work to be equal to that.
$endgroup$
– Acccumulation
Jan 3 at 20:01












$begingroup$
@Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 20:10




$begingroup$
@Accumulation I see, so because the area relates to position (in this graph) from it I could obtain the point from which it nulls that right?. I totally overlooked the fact that acceleration was changing. Still I'm confused what is meant by part b in the question, how do I know that $x=0$ the block is moving to the right, does it mean that passed that point the block will move to the left and then to the right?.
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 20:10












$begingroup$
@Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 20:38




$begingroup$
@Accumulation. By following your hint I obtained an expression like this $frac{-40-a}{2}b=-25$ being $a$ and $b$ the sides of the trapezoid. But to choose which?. There are countless possibilities from the graph, however if I reduce it to the most visual being the one at $a=F=-20$, then $b=x=frac{5}{6}$ this doesn't check. Now if I try$a=-10$then$b=1$which correspond to the "answer". However,how about the point passed$x=1$the block moves to the left,negative work. But is part (b) where I'm stuck at now,how to relate it? I thought that it was at$x=1.66,m$where the area counters the left side
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 20:38












$begingroup$
@ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
$endgroup$
– amd
Jan 3 at 21:13




$begingroup$
@ChrisSteinbeckBell Why would you choose anything but $a=0$? That’s where the retarding force is first applied, after all.
$endgroup$
– amd
Jan 3 at 21:13











1












$begingroup$

Hint.



You have



$$
m ddot x = f(x) = begin{cases}
5 (6 x-8) & x le 2 \
20 & 2lt x le 3 \
10 (2 x-4) & x gt 3
end{cases}
$$



then



$$
mdot x ddot x = f(x)dot xto frac m2 frac{d}{dt}dot x ^2 = frac{d}{dt}F(x)
$$



then integrating



$$
frac 12 m (v^2-v_0^2) = F(x) = begin{cases}
15 x^2-40 x & xleq 2 \
20 x-60 & 2<xleq 3 \
20 left(frac{x^2}{2}-2 xright)+30 & 3 < x
end{cases}
$$



Attached the plot for $f(x)$ in blue, the plot for $F(x)$ in red and the plot for $v(x)$ in black showing the points in which $v = v_0$ which occurs at $x = 0$ and $x = 3$



enter image description hereenter image description hereenter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:57










  • $begingroup$
    I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 21:04










  • $begingroup$
    @ChrisSteinbeckBell I hope now helps more.
    $endgroup$
    – Cesareo
    Jan 3 at 21:06










  • $begingroup$
    @ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
    $endgroup$
    – Cesareo
    Jan 3 at 21:08






  • 1




    $begingroup$
    @ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
    $endgroup$
    – Andrei
    Jan 4 at 17:02
















1












$begingroup$

Hint.



You have



$$
m ddot x = f(x) = begin{cases}
5 (6 x-8) & x le 2 \
20 & 2lt x le 3 \
10 (2 x-4) & x gt 3
end{cases}
$$



then



$$
mdot x ddot x = f(x)dot xto frac m2 frac{d}{dt}dot x ^2 = frac{d}{dt}F(x)
$$



then integrating



$$
frac 12 m (v^2-v_0^2) = F(x) = begin{cases}
15 x^2-40 x & xleq 2 \
20 x-60 & 2<xleq 3 \
20 left(frac{x^2}{2}-2 xright)+30 & 3 < x
end{cases}
$$



Attached the plot for $f(x)$ in blue, the plot for $F(x)$ in red and the plot for $v(x)$ in black showing the points in which $v = v_0$ which occurs at $x = 0$ and $x = 3$



enter image description hereenter image description hereenter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:57










  • $begingroup$
    I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 21:04










  • $begingroup$
    @ChrisSteinbeckBell I hope now helps more.
    $endgroup$
    – Cesareo
    Jan 3 at 21:06










  • $begingroup$
    @ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
    $endgroup$
    – Cesareo
    Jan 3 at 21:08






  • 1




    $begingroup$
    @ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
    $endgroup$
    – Andrei
    Jan 4 at 17:02














1












1








1





$begingroup$

Hint.



You have



$$
m ddot x = f(x) = begin{cases}
5 (6 x-8) & x le 2 \
20 & 2lt x le 3 \
10 (2 x-4) & x gt 3
end{cases}
$$



then



$$
mdot x ddot x = f(x)dot xto frac m2 frac{d}{dt}dot x ^2 = frac{d}{dt}F(x)
$$



then integrating



$$
frac 12 m (v^2-v_0^2) = F(x) = begin{cases}
15 x^2-40 x & xleq 2 \
20 x-60 & 2<xleq 3 \
20 left(frac{x^2}{2}-2 xright)+30 & 3 < x
end{cases}
$$



Attached the plot for $f(x)$ in blue, the plot for $F(x)$ in red and the plot for $v(x)$ in black showing the points in which $v = v_0$ which occurs at $x = 0$ and $x = 3$



enter image description hereenter image description hereenter image description here






share|cite|improve this answer











$endgroup$



Hint.



You have



$$
m ddot x = f(x) = begin{cases}
5 (6 x-8) & x le 2 \
20 & 2lt x le 3 \
10 (2 x-4) & x gt 3
end{cases}
$$



then



$$
mdot x ddot x = f(x)dot xto frac m2 frac{d}{dt}dot x ^2 = frac{d}{dt}F(x)
$$



then integrating



$$
frac 12 m (v^2-v_0^2) = F(x) = begin{cases}
15 x^2-40 x & xleq 2 \
20 x-60 & 2<xleq 3 \
20 left(frac{x^2}{2}-2 xright)+30 & 3 < x
end{cases}
$$



Attached the plot for $f(x)$ in blue, the plot for $F(x)$ in red and the plot for $v(x)$ in black showing the points in which $v = v_0$ which occurs at $x = 0$ and $x = 3$



enter image description hereenter image description hereenter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 22:51

























answered Jan 3 at 19:06









CesareoCesareo

9,6223517




9,6223517












  • $begingroup$
    I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:57










  • $begingroup$
    I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 21:04










  • $begingroup$
    @ChrisSteinbeckBell I hope now helps more.
    $endgroup$
    – Cesareo
    Jan 3 at 21:06










  • $begingroup$
    @ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
    $endgroup$
    – Cesareo
    Jan 3 at 21:08






  • 1




    $begingroup$
    @ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
    $endgroup$
    – Andrei
    Jan 4 at 17:02


















  • $begingroup$
    I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 19:57










  • $begingroup$
    I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
    $endgroup$
    – Chris Steinbeck Bell
    Jan 3 at 21:04










  • $begingroup$
    @ChrisSteinbeckBell I hope now helps more.
    $endgroup$
    – Cesareo
    Jan 3 at 21:06










  • $begingroup$
    @ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
    $endgroup$
    – Cesareo
    Jan 3 at 21:08






  • 1




    $begingroup$
    @ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
    $endgroup$
    – Andrei
    Jan 4 at 17:02
















$begingroup$
I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 19:57




$begingroup$
I was hoping that the answer would rely into precalculus anyways... $v_{0}$ is known so what's with $v$ do what?. $v=0$ and from then on to calculate from $F(x)$ the position, but what $F(x)$ should I use?. Let's say between $[0,2]$ with $F(x)=30x-40$ I obtained $x=frac{1}{2}$ and again this is not close to the answer. Can you further help? I'm stuck.
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 19:57












$begingroup$
I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 21:04




$begingroup$
I think in this case a much better help than a plot would be by stating which is the part where I'm stuck on with the calculation. The second graph is I don't understand, does it states explicitly the values of $x$? What's the horizontal line for, to indicate just $25$?.
$endgroup$
– Chris Steinbeck Bell
Jan 3 at 21:04












$begingroup$
@ChrisSteinbeckBell I hope now helps more.
$endgroup$
– Cesareo
Jan 3 at 21:06




$begingroup$
@ChrisSteinbeckBell I hope now helps more.
$endgroup$
– Cesareo
Jan 3 at 21:06












$begingroup$
@ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
$endgroup$
– Cesareo
Jan 3 at 21:08




$begingroup$
@ChrisSteinbeckBell The equation you should use is $mddot x = f(x)$ with all it's consequences.
$endgroup$
– Cesareo
Jan 3 at 21:08




1




1




$begingroup$
@ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
$endgroup$
– Andrei
Jan 4 at 17:02




$begingroup$
@ChrisSteinbeckBell Just a quick note about your pre-calculus request: If you look at the last equation, what you need to notice is that on the left is the change in kinetic energy, and on the right is the work done by the force. $|Delta E_{kinetic}=W$. You can get this work from simple geometry, since it's the area under the curve of the force vs distance. And all you have is triangles and rectangles.
$endgroup$
– Andrei
Jan 4 at 17:02


















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