Work Against Gravity: Understanding Force and Energy

[jldibble]

I'm figuring that this has been asked before, but I couldn't locate a previous thread.

Here's my problem:

A 1.0 kg ball is lifted at a constant speed to a height of 1.0 meter above the ground. The work done to the ball against gravity is about 10 joules.

The same 1.0 kg ball is lifted with a force of 15 Newtons to a height of 1.0 meter above the ground. The work done to the ball in this case is 15 joules.


What is getting me confused is that the work done to lift the ball 1.0 meter should be the same no matter what. And the work energy theorem says that a 1.0 kg object 1.0 meter above the ground should have a GPE of 10 joules. So my problem is with lifting objects with a force greater than their weight.

What am I getting mixed up here?

 

[Doc Al]

A 1.0 kg ball is lifted at a constant speed to a height of 1.0 meter above the ground. The work done to the ball against gravity is about 10 joules.

That's the minimum amount of work you need to do to just lift the ball against gravity.

The same 1.0 kg ball is lifted with a force of 15 Newtons to a height of 1.0 meter above the ground. The work done to the ball in this case is 15 joules.

Here you did more work that necessary to lift the ball. So that extra work goes into the increased KE of the ball.

What is getting me confused is that the work done to lift the ball 1.0 meter should be the same no matter what.

No, not really.

And the work energy theorem says that a 1.0 kg object 1.0 meter above the ground should have a GPE of 10 joules.

No, the work energy theorem just says that the work you do must equal the change in energy of the object. And it does!

 

[Dale]

You are correct, the 15 N force will result in 15 J of work done. At the end the ball will have 10 J of GPE and 5 J of KE.

 

[jldibble]

Oh, of course... Easy enough. Thanks!

 

[PJC01]

I can offer some clarification on this topic. The work done on an object is defined as the force applied multiplied by the distance moved in the direction of the force. In this case, the work done against gravity is equal to the force of gravity (weight) multiplied by the distance the object is lifted. This is why the work done in the first scenario, where the ball is lifted at a constant speed, is 10 joules (weight of 1 kg x 1 meter = 10 joules).

In the second scenario, where a force of 15 Newtons is applied to lift the ball, the work done is 15 joules (15 Newtons x 1 meter = 15 joules). This is because the force applied is greater than the weight of the object, resulting in a greater work done.

It is important to note that the work done against gravity is not the same as the potential energy (GPE) of the object. The GPE of an object is the energy it possesses due to its position in a gravitational field, and is calculated using the formula mgh (mass x gravity x height). In this case, the GPE of the 1 kg ball lifted to a height of 1 meter would be 9.8 joules, which is equal to the work done against gravity in the first scenario.

In summary, the work done against gravity is dependent on the force applied and the distance moved, while the GPE of an object is dependent on its mass, height, and the strength of the gravitational field. I hope this helps clarify any confusion you may have had.