You raise a rock of mass M a distance h at a constant velocity. Neglect friction.

The change in the potential energy (final minus initial) equals

Question

You raise a rock of mass M a distance h at a constant velocity. Neglect friction.

The change in the potential energy (final minus initial) equals

BlackTom AI · Accepted Answer

Question restatement: You raise a rock of mass M a distance h at a constant velocity. Neglect friction. The change in the potential energy (final minus initial) equals

Answer options:
1) the negative of the change in the kinetic energy of the rock.
2) the work done on the rock by the force of gravity.
3) zero, since the velocity of the rock does not change.
4) the work done on the rock by you against the force of gravity.

Now, let's analyze each option:

Option 1: the negative of the change in the kinetic energy of the rock.
- By the work-energy theorem, the net work equals the change in kinetic energy: W_net = ΔK. Since the velocity is constant, ΔK = 0 and W_net = 0. The change in potential energy ΔU is not generally equal to the negative of ΔK; rather, for conservative forces like gravity, ΔU = -W_gravity. This option incorrectly equates ΔU with -ΔK in a situation where gravity is involved but not describing the correct energy relationship, so this is not correct.

Option 2: the work done on the rock by the force of gravity.
- Gravity is a conservative force that decreases potential energy when you raise the mass. The work done by gravity during the raise is negative and equals -Mg h. The change in potential energy ΔU = U_final − U_initial equals +Mg h, which is the negative of the work done by gravity. Therefore, this option does not match the standard definition ΔU = -W_gravity.

Option 3: zero, since the velocity of the rock does not change.
- While ΔK = 0 due to constant velocity, the potential energy change ΔU is not zero when the height increases; gravitational potential energy increases by Mg h. This option confuses kinetic energy behavior with potential energy change and is incorrect.

Option 4: the work done on the rock by you against the force of gravity.
- When you lift the rock at constant speed, you perform positive work against gravity equal in magnitude to Mg h. This work increases the rock’s gravitational potential energy by ΔU = Mg h. This matches the standard relationship for a conservative force: ΔU = -W_gravity, and W_you_against_gravity = +Mg h = ΔU. The physical interpretation is that your lifting work converts chemical or other energy into gravitational potential energy. Therefore, this option correctly identifies the change in potential energy as the work you perform against gravity.

In summary, option 4 aligns with the physics of gravitational potential energy for a mass raised by an external agent at constant velocity, while the other options either misstate the energy relationships or rely on incorrect assumptions about motion or forces involved.

You raise a rock of mass M a distance h at a constant velocity. Neglect friction. The change in the potential energy (final minus initial) equals单项选择题

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