Consider an object which has been given a speed $v$ on a rough horizontal surface. As time passes, the object covers a distance $l$ until it stops because of friction. Now,
Initial kinetic energy = $\frac{1}2mv^2$
And final kinetic energy is zero. Therefore, work done by friction on the object is equal in magnitude to $\frac{1}2mv^2$.
Now here is the part that I found weird: Consider another frame moving with a speed $v_0$ in the same direction with respect to the ground frame. Now, kinetic energy of the original object with respect to this new frame is $\frac{1}2m(v-v_0)^2$.
And, the final kinetic energy is equal to $\frac{1}2mv_0^2$.
So this means that the work done by frictional force, in this case, will have a magnitude of $\frac{1}2m[(v-v_0)^2-v_0^2]$, which is obviously different from the value which we get with respect to a stationary frame.
And this part seems very unintuitive to me. How is it possible for the same force to do different amounts of work in two different inertial frames? (I would consider it unintuitive even if we consider non inertial frames, after considering pseudo forces).
And if we were to do more calculations based on the two values of the work done by friction, we would land on different values of some quantities which aren't supposed to be different in any frame. For example, the coefficient of friction would be different, as the amount of frictional force is constant, acting over a distance $l$. We can say that Work done by frictional force is $\alpha$$mgl$, where $\alpha$ is the coefficient of friction and $g$ is the acceleration due to gravity. We can clearly see that $\alpha$$mgl$ equals two different values.
So, is this just how physics works, or is there something wrong here?
Best Answer
You have correctly discovered that power, work, and kinetic energy are all frame variant. This is well-known for centuries, but is always surprising to a student when they first discover it. For some reason, it is not part of a standard physics curriculum.
So, the reason that this is disturbing to every student who encounters it is that it seems irreconcilable with the conservation of energy. If the work done is different in different reference frames then how can energy be conserved in all frames?
The key is to recognize that the force doing work acts on two bodies. In this case the object and the horizontal surface. You must include both bodies to get a complete picture of the conservation of energy.
Consider the situation in your example from an arbitrary frame where the horizontal surface (hereafter the "ground") is moving at a velocity $u$, the ground frame then being the frame $u=0$. Let the ground have mass $M$. The initial kinetic energies are:
$$KE_{obj}(0)=\frac{1}{2}m (v+u)^2$$ $$KE_{gnd}(0)=\frac{1}{2} M u^2$$
Now, the friction force $-f$ acts on the object until $v_{obj}(t_f)=v_{gnd}(t_f)$. Solving for the time gives $$t_f=\frac{m M v}{(m+M) f}$$ and, by Newton's 3rd law, a force $f$ acts on the ground for the same time.
At $t_f$ the final kinetic energies are:
$$KE_{obj}(t_f)=\frac{1}{2} m \left(\frac{Mu+m(u+v)}{m+M} \right)^2$$ $$KE_{gnd}(t_f)=\frac{1}{2} M \left(\frac{Mu+m(u+v)}{m+M} \right)^2$$ so $$\Delta KE_{obj}+\Delta KE_{gnd}=-\frac{m M v^2}{2(m+M)}$$
Note importantly that the total change in KE is independent of $u$, meaning that it is frame invariant. This is the amount of energy that is converted to heat at the interface. So even though the change in KE for the object itself is frame variant, when you also include the ground then you find that the total change in kinetic energy is frame invariant which allows energy to be conserved since the amount of heat generated is frame invariant.