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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?

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  • $\begingroup$ This may not help you see why it's incorrect, but to see that it's incorrect, I think it's helpful to notice that the rock is being accelerated upward, so its kinetic energy and potential energy are both increasing -- unlike in a "falling" scenario, accelerating downward, where potential energy is decreasing because it's being converted to kinetic energy. $\endgroup$ – ruakh Apr 30 '17 at 19:56
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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.

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The scenario in the question and the scenario of the falling rock are not time inversions of each other:

  • The falling rock starts with zero velocity. The lifted rock ends with some non-zero velocity.

  • The falling rock ends with some non-zero velocity. The lifted rock starts with zero velocity.

  • The lifted rock is lifted by some external force (going against gravitational force). The falling rock is accelerated by gravitational force.

The time-inverted scenario to the falling rock would be a rock that receives a kick of kinetic energy at the beginning of the time interval and flies upwards until it reaches its apex – when all of the initial kinetic energy has been translated to potential energy. By contrast, in the scenario in question, the lifted rock receives kinetic energy all the time.

On the other hand, to obtain a potential energy equivalent to the kinetic energy in question, you would have to:

  1. switch off the force after the time in question;
  2. let the rock fly further upwards, thus translating the kinetic energy in question to potential energy;
  3. wait till the rock reaches its apex;
  4. take the difference between the potential energy at the apex and the time point when the force was switched off.
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When a rock is falling from a given height, the only forces on it is the gravitational force which is a conservative force. The concept of potential energy comes only when all the forces acting on the body are conservative. In your question it's not given if the force $F$ is conservative or not. If $F$ is also conservative, then yes option $D$ is also correct. But I think maybe the question meant to say gravitational potential energy instead of potential energy. If that's the case, then option $D$ is not correct because the change in gravitational potential energy only corresponds to the work done by $W$. So, the work done by $F$ is not taken into account when you calculate the change in gravitational potential energy.

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Answer D could only be correct if energy was conserved. But energy isn't conserved because the force $F$ is an external force that's doing work on the system and putting energy into it.

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What is different about the scenario presented in the problem that doesn’t make D correct?

The rock is gaining, not losing potential energy. The correct formulation in terms of potential energy would be;

Work done by F minus the change in potential energy.

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The difference here is that the block is being pushed UP not DOWN


The potential energy of a body increases with the height from the centre of the earth. enter image description here The above image shows roughly the potential energy levels for a body of mass 0.5 kg. Let us analyse 2 different cases:

  1. Body falls from A to B:
    Here, the potential energy of the body changes by $\delta PE = -10 J$. But since the potential energy decreases, it must be converted to kinetic energy. Thus here, the change in kinetic energy is the magnitude of change of potential energy
  2. Body moves from B to A under the application of force $F$:
    Here, the change in potential energy is +10 J. Since the change in potential energy is positive, some other external force must do work on the body. In our situation, that external force is the force $F$, which does a net work of $F*(4-2)=2F$. Now, we can write F as $W+(F-W)$ where $W$ is the weight of the body. Thus the net work done by F is $2W+2(F-W)$. But $2W$ is the change in the potential energy, which implies that the remainging work done must be used to increase the kinetic energy.
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protected by Qmechanic Apr 30 '17 at 11:16

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