Work due to Gravity problem

[Reverie29]

I feel like I'm missing something, specifically in regard to the total amount of work. Can someone set me straight?

Homework Statement

Suppose you lift a heavy weight from the floor to high over your head and let it fall back to the floor. An observer says you did no work because it wound up in its original position. You disagree. Who’s right?

Homework Equations

W = F*d

The Attempt at a Solution

Work is equal to force * distance, or in other words, the amount of force exerted over a certain distance. Work is also a scalar quantity, which makes the direction irrelevant to the problem (there are no negative or positive signs indicating the direction). All scalars are treated as positive values. Although the weight winds up in the same position, work is still done. Therefore, the total amount of work done is equal to the weight of the weight multiplied by the distance it is lifted up, in addition to when it is allowed to fall back to the floor. However, when it is lifted up, the person lifting the object is doing work. When it is left to fall, the person is no longer doing work and gravity does work pulling the object back to the ground.

 

[alphysicist]

Hi Reverie29,

I feel like I'm missing something, specifically in regard to the total amount of work. Can someone set me straight?

Homework Statement

Suppose you lift a heavy weight from the floor to high over your head and let it fall back to the floor. An observer says you did no work because it wound up in its original position. You disagree. Who’s right?

Homework Equations

W = F*d

The Attempt at a Solution

Work is equal to force * distance, or in other words, the amount of force exerted over a certain distance. Work is also a scalar quantity, which makes the direction irrelevant to the problem (there are no negative or positive signs indicating the direction).

The directions of the force and displacement are very important, as work can be a negative quantity. For constant forces, the work done by a force F is

$$W=Fd\cos\theta$$

where $\theta$ is the angle between the directions of F and d.

All scalars are treated as positive values. Although the weight winds up in the same position, work is still done. Therefore, the total amount of work done is equal to the weight of the weight multiplied by the distance it is lifted up, in addition to when it is allowed to fall back to the floor. However, when it is lifted up, the person lifting the object is doing work. When it is left to fall, the person is no longer doing work and gravity does work pulling the object back to the ground.

The question only asked about the work that you did on the box, and I believe you got that correct.

But to add a bit to what you have written: since the block has zero kinetic energy at the beginning and ending of the motion, the total amount of work done on the block is zero. You are right that the person does positive work on the box while lifting it up, and gravity does positive work on the box while the box falls. So what else does work on the box, to make the total work be zero?

 

[baba89]

Therefore, the total amount of work is not just the work done by the person lifting the object, but also the work done by gravity pulling it back down.

I agree with your reasoning. It is important to note that work is not just the physical act of lifting an object, but also the work done by other forces, such as gravity, in this case. The total amount of work done is the sum of the work done by the person and the work done by gravity. So, both you and the observer are partially correct. The person lifting the object is doing work and so is the force of gravity. This concept is known as the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy. In this case, the object starts with no kinetic energy and ends with no kinetic energy, but the work done by the person and gravity still contribute to the total amount of work done.