Physics Asked by k.ali on March 31, 2021
I have a quite a simple question that is escaping me right now. The work done on any body is just the net force multiplied by the distance the force acts on it for.
For example: the work being done on a lawnmower moving some distance d taking into account the applied force and the frictional force is: (Fapplied – Ffriction)*d.
Why doesn’t the same apply when looking at stress strain curves? The force described by any point on the F = kx curve is the force applied. However, since the strain puts an equivalent force back, the net force at each point during elongation is not equal to kx. Since kx is not the net force, why do we consider the area under the kx curve to be the energy stored in the object.
Thanks
For example: the work being done on a lawnmower moving some distance d taking into account the applied force and the frictional force is: (Fapplied - Ffriction)*d.
If $F_{applied} - F_{friction} > 0$ the lawnmower is accelerated. Normally it is not the case, and still we do work. That means: when we say work, it is necessary to specify which force we select. In the case of the lawnmower it is our force. In the case of the stress-strain curve it is the elastic force $F = kx$.
Correct answer by Claudio Saspinski on March 31, 2021
The work done on a body is not the net force multiplied by the distance the force acts for. The work done on a body is the sum of the work done by the separate forces acting on the body. Different forces may act over different distances, so you cannot simply multiple the net force by one single distance.
Two examples:
(A) A massless spring with spring constant $k$ is fixed at one end and compressed by a distance $d$ at the other end. The net force on the spring is always zero - at any time the force at one end is balanced by an equal and opposite force at the other end. But the work done on the spring is not zero. The force at the fixed end does not move, so it does no work. So the work done on the spring (and hence the energy stored in the spring) is $int_0^d kx dx = frac 1 2 kd^2$.
(B) Two children push a roundabout with radius $r$ through an angle $theta$ radians. The children both push clockwise with the same constant tangential force $F$ but at opposite sides of the roundabout. Once again, the net force on the roundabout is always zero, but the total work done on the roundabout is not zero. Each child pushes for a distance $r theta$ so the work done on the roundabout is $2Frtheta$.
Answered by gandalf61 on March 31, 2021
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