A rock is lifted for a certain time by a force $F$ that is greater in magnitude than the rock’s weight $W$. The change in kinetic energy of the rock during this time is equal to the
A. work done by the net force ($F – W$)
B. work done by $F$ alone
C. work done by $W$ alone
D. difference in the potential energy of the rock before and after this time.
The correct answer is A. I understand why this is so: The work–energy theorem states that the net work done on an object is equal to its change in kinetic energy.
What I’m having trouble understanding is why answer choice D is incorrect. If a rock is falling from a given height, the rock’s change in kinetic energy (which goes from $0$ to $\frac{1}{2}mv^2$) is equal to its change in potential energy (which goes from $mgh$ to $0$). What is different about the scenario presented in the problem that doesn’t make D correct?
Best Answer
In this question the rock is the system which you are asked to consider and the rock by itself does not have gravitational potential energy.
It is the rock and Earth system which possesses gravitational potential energy.
When a rock falls and you are treating the rock as the system the gravitational force of attraction on the rock by the Earth is an external force which is doing work on the rock.
Having the rock and the Earth as the isolated system there are no external forces acting on the system.
The gain in kinetic energy of the system (rock and Earth) is equal to the loss if gravitational potential energy of the system.
The gain in kinetic energy of the Earth is usually neglected because it is so small compared with the gain in kinetic energy of the rock.
In the case where a force $F$ is lifting the rock by a distance $h$ the two external forces on the rock (the system) are the force $F$ and the weight of the rock $W$.
The net work done on the rock is $(F-W)h$ and this is equal to the gain in the kinetic energy of the rock $\Delta E_{\rm K}$
If the rock and the Earth is the system the only external force is $F$.
That single force $F$ cannot move the rock further away from the Earth.
All that force can do is move the centre of mass of the Earth and rock system.
To separate the rock and the Earth two forces of magnitude $F$ but opposite in direction must act on the Earth and the rock.
Because the mass of the Earth is so much greater than the mass of the rock the displacement of the Earth relative to the centre of mass of the Earth and rock system is very small compared to the displacement of the rock.
The work done on the Earth by the external forces is negligible compared with the work done on the rock.
The work done by the external forces on the Earth and rock system is $Fh$.
The increase in gravitational potential energy of the system is $Wh$.
$\Rightarrow Fh = Wh + \Delta E_{\rm K}$ which is equivalent to $(F-W)h = \Delta E_{\rm K}$ the result obtained when considering the system as the rock alone.