# Work calculations

1. Aug 22, 2010

### hydrogen1

Hey guys, I understand that if you lift an object a short distance in the air that it takes work because you are moving against the the weight of the object through a distance. So the equation in my mind is W=mgdcos$$\Theta$$ where $$\Theta$$=0. But what I can't figure out is what if I lift the object with a force that accelerates the object F>Weight. If I lift an object with a force equal to it's weight is the work different than if I accelerate the object to the same height? I hope this makes sense.

2. Aug 22, 2010

### Andy Resnick

No, because the work is not dependent on the path the object took. The work (in this case) is only given by the change in gravitational potential energy of the lifted object, and the potential energy is only due to the position of the object, not how fast it's moving at that position.

Now, if you want to talk about the work your muscles expended during the lifting, that's most likely different depending on the speed of lifting. Most likely, a machine could also consume different amounts of power depending on the rate of lifting.

3. Aug 22, 2010

### hydrogen1

Wouldn't I be putting in the work to change potential energy and giving the object kinetic energy too? I do understand that the potential energy is a state function

4. Aug 23, 2010

### Andy Resnick

I think I understand your question a little better. Work = Force*distance, and if you vary the force (i.e. the object's acceleration), you can vary the work done on the object. Some of that work will be used to change the potential energy, and the rest goes to kinetic energy.

5. Aug 23, 2010

### georgir

Totally not so. The work is given by the change in total energy. That means potential energy change, but also kinetic energy change. If the object has more speed at the end of the exercise, then you did more work, even if it is at the same height so far. This additional speed means that eventually the object will continue moving up and reach higher height on its own.

Work is directly multiplication of the magnitude of the force by the distance traveled in the same direction as the force. So obviously lifting it with more force does more work.

6. Aug 23, 2010

### hydrogen1

Ok, thanks this is now clear to me. The force applied - the weight of the object goes to kinetic energy

7. Aug 23, 2010

### Andy Resnick

I had originally thought the OP meant the object is at rest at the two endpoints. When the OP clarified, my answer was adjusted.

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