If you were to lift an object above the ground at a constant speed, you would have to exert a force of equal magnitude and opposite the direction of gravity onto it. However, wouldn't these forces cancel out and thus not affect the object's position at all? Also, if this object was elevated at a constant speed, I understand that it would gain in gravitational potential energy since its height above the ground is increasing. But what about its kinetic energy? Although there is no net force in the direction of the object's motion (constant speed, so equilibrium), the object is still moving, so wouldn't it possess some amount of kinetic energy? If so, what becomes of it once the object stops at some height, h?

  • $\begingroup$ The ground was already exerting an equal and opposite force on the object. You lifting it already provided the unbalanced force that lead to acceleration. $\endgroup$ – Smartybartfast Aug 25 '16 at 4:43

To understand why the answer they give in the physics text book makes sense, there's one slightly counter-intuitive thing to remember: if you try to lift something up "at a constant speed" the force you apply is not constant. We humans lift objects at a constant speed so intuitively, that it is easy for us to get the illusion that we're applying a constant force, but that's not actually the case.

If an object is stationary, and you apply a force exactly equal to the force of gravity but in the upwards direction, the object will not move. This should make sense, if you think about it. The "normal force" from the ground holding up the object is exactly equal to the force of gravity if the object is lying there on the ground. To lift it off the ground, you have to apply more upward force than the force of gravity so that there is a net upward force.

Now while you're doing this, the object is not moving at a constant speed. It is accelerating, exactly as predicted by F=ma. After a short while, its upward speed is exactly the "constant speed" you wanted, and you intuitively decrease the force you are exerting on the box until the force you are applying matches that of gravity exactly. In most intuitive cases, this acceleration occurs quickly, so we get tempted to ignore it, but it's absolutely there. In other situations, it's not so easily ignored. Consider the case of large cranes:

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When you have a crane lifting something as massive as this, it turns out that it's not effective to have the operator's joystick controlling the position of the crane up and down. They actually control the force that the crane applies when lifting the object. They have to be aware of the acceleration phases that we often intuitively overlook. They will actually put more force on the object to get it moving upward and then decrease their force so that it doesn't accelerate up out of control.

When you reach your height, h, you stop. To stop, what you actually do is decrease your force to less than that of gravity, and it decelerates. Like before, the process is not instantaneous, but on human scales it often occurs fast enough that we don't pay attention to it. Once it has stopped moving upward, we once again apply a force exactly equal to that of gravity to keep it stationary.

So during the initial acceleration period, your object is gaining kinetic energy because it velocity is increasing. During the middle period, where your force matches gravity's exactly, its velocity stays the same, so its kinetic energy remains the same. Any work you are doing (force * distance) is going into increasing its potential energy. Near the top, the object decelerates back to motionless. During this time, all of the kinetic energy you built up during the acceleration period is converted into potential energy (along with any potential energy coming from the fact that you're still applying force while its decelerating, it's just less force than before)

  • $\begingroup$ Very concise answer. I thought that there must be a net force to produce some initial acceleration, but all of the situations which I've studied had no mention of it. The crane example was exactly what was bothering me, since such an acceleration cannot be ignored. Thanks for clarifying. $\endgroup$ – Tether Current Aug 25 '16 at 19:59

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