what is the physical interpretation of a work done by a vector field F = zero?
If we have a field vector $vec F= (e^x sin y )hat i +(e^x cos y) hat j$ and there is a particle that has moved along the path from $(0, 0)$ to $(1, pi/2)$ due to this field. If we calculate the work done by this vector field , we will get work$=0$. What is the physical meaning of this value ? The particle has moved due to work , but how this work is equal to zero ?
vector-fields
edited Nov 24 at 21:07
Andrei
10.8k21025
asked Nov 24 at 20:49
John adams
206
add a comment |
If we have a field vector $vec F= (e^x sin y )hat i +(e^x cos y) hat j$ and there is a particle that has moved along the path from $(0, 0)$ to $(1, pi/2)$ due to this field. If we calculate the work done by this vector field , we will get work$=0$. What is the physical meaning of this value ? The particle has moved due to work , but how this work is equal to zero ?
vector-fields
edited Nov 24 at 21:07
Andrei
10.8k21025
asked Nov 24 at 20:49
John adams
206
I found the work was equal to $e$, not $0$.
– Syuizen
Nov 24 at 21:03
Imagine a perfect electric car driving over hilly terrain. If the total work is zero, that might that it’s regenerated as much energy going down hills as it expended going up.
– amd
Nov 24 at 22:06
add a comment |
If we have a field vector $vec F= (e^x sin y )hat i +(e^x cos y) hat j$ and there is a particle that has moved along the path from $(0, 0)$ to $(1, pi/2)$ due to this field. If we calculate the work done by this vector field , we will get work$=0$. What is the physical meaning of this value ? The particle has moved due to work , but how this work is equal to zero ?
vector-fields
edited Nov 24 at 21:07
Andrei
10.8k21025
asked Nov 24 at 20:49
John adams
206
If we have a field vector $vec F= (e^x sin y )hat i +(e^x cos y) hat j$ and there is a particle that has moved along the path from $(0, 0)$ to $(1, pi/2)$ due to this field. If we calculate the work done by this vector field , we will get work$=0$. What is the physical meaning of this value ? The particle has moved due to work , but how this work is equal to zero ?
vector-fields
vector-fields
edited Nov 24 at 21:07
Andrei
10.8k21025
asked Nov 24 at 20:49
John adams
206
edited Nov 24 at 21:07
Andrei
10.8k21025
asked Nov 24 at 20:49
John adams
206
edited Nov 24 at 21:07
Andrei
10.8k21025
edited Nov 24 at 21:07
Andrei
10.8k21025
edited Nov 24 at 21:07
Andrei
10.8k21025
10.8k21025
asked Nov 24 at 20:49
John adams
206
asked Nov 24 at 20:49
John adams
206
asked Nov 24 at 20:49
John adams
206
206
I found the work was equal to $e$, not $0$.
– Syuizen
Nov 24 at 21:03
Imagine a perfect electric car driving over hilly terrain. If the total work is zero, that might that it’s regenerated as much energy going down hills as it expended going up.
– amd
Nov 24 at 22:06
add a comment |
I found the work was equal to $e$, not $0$.
– Syuizen
Nov 24 at 21:03
Imagine a perfect electric car driving over hilly terrain. If the total work is zero, that might that it’s regenerated as much energy going down hills as it expended going up.
– amd
Nov 24 at 22:06
I found the work was equal to $e$, not $0$.
– Syuizen
Nov 24 at 21:03
I found the work was equal to $e$, not $0$.
– Syuizen
Nov 24 at 21:03
Imagine a perfect electric car driving over hilly terrain. If the total work is zero, that might that it’s regenerated as much energy going down hills as it expended going up.
– amd
Nov 24 at 22:06
Imagine a perfect electric car driving over hilly terrain. If the total work is zero, that might that it’s regenerated as much energy going down hills as it expended going up.
– amd
Nov 24 at 22:06
add a comment |
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I found the work was equal to $e$, not $0$.
– Syuizen
Nov 24 at 21:03
Imagine a perfect electric car driving over hilly terrain. If the total work is zero, that might that it’s regenerated as much energy going down hills as it expended going up.
– amd
Nov 24 at 22:06