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# Can't understand Work Done concept Watch

1. So I learnt that work done can be summarised as Fs where s is in the forces direction and I thought I understood it. Yet when given this simple question:
A 2kg particle falls from rest 10m with no air resistance. What is the total work done on the particle?
I did weight x 10 to get 196J but the answer was actually 0J.
I asked a friend and he couldn't really explain fully but said the reason for this is I should of calculated the the Work Done Against Resistance and as there is no resistance, the answer is 0.
Can anyone further explain this? What's the difference between work done and work done against resistance? How do I know if a questions asking for work done against resistance? Is there a difference between work done BY a particle and work done ON a particle? I'm just really confused.
2. Work done is equal to force multiplied by displacement: W = Fs
Work done ON the particle means work done against the particle while it is falling, by resistive forces - so if there were resistive forces like air resistance, air resistance would do work against the particle to reduce it's terminal velocity. But since there are no resistive forces, there is nothing to oppose the falling particle, therefore no work can be done on/against the particle.
3. I'm inclined to agree with 196J. The force of gravity has done work on the mass - it's kinetic energy has changed.
4. (Original post by ghostwalker)
I'm inclined to agree with 196J. The force of gravity has done work on the mass - it's kinetic energy has changed.
I agree with you.
5. (Original post by Bobrey)
So I learnt that work done can be summarised as Fs where s is in the forces direction and I thought I understood it. Yet when given this simple question:
A 2kg particle falls from rest 10m with no air resistance. What is the total work done on the particle?
I did weight x 10 to get 196J but the answer was actually 0J.
I asked a friend and he couldn't really explain fully but said the reason for this is I should of calculated the the Work Done Against Resistance and as there is no resistance, the answer is 0.
Can anyone further explain this? What's the difference between work done and work done against resistance? How do I know if a questions asking for work done against resistance? Is there a difference between work done BY a particle and work done ON a particle? I'm just really confused.
(Original post by ghostwalker)
I'm inclined to agree with 196J. The force of gravity has done work on the mass - it's kinetic energy has changed.
(Original post by atsruser)
I agree with you.

Me too. Sloppily worded question.
6. (Original post by Bobrey)
A 2kg particle falls from rest 10m with no air resistance. What is the total work done on the particle?
I did weight x 10 to get 196J but the answer was actually 0J.

What's the difference between work done and work done against resistance?
How do I know if a questions asking for work done against resistance? Is there a difference between work done BY a particle and work done ON a particle?
I'd argue that there's conservation of energy. The gravitational potential energy decreases from 196J to 0J while the kinetic energy increases from 0J to 196J then 0J when it hits the ground. A gain of 196J of EK is balanced by the loss of 196J of GPE, hence the total/net work done is zero.

Work done is just the energy required to move an object through the action of a force (in the direction of the force). Work done against resistance is the Fxd required to move an object through an opposing force.

Work done ON a particle is greater than 0.
Work done BY a particle is less than 0. It's essentially a result of Newton's third law (equal and opposite forces). Energy is a scalar, so a negative sign not to do with direction, it's just a relative energy.

Say you push a box forwards. The work you do on the box might be 10J, and the work done by the box on you is -10J.
7. (Original post by Laurasaur)
I'd argue that there's conservation of energy.
That's not really going out on a limb, is it?

The gravitational potential energy decreases from 196J to 0J while the kinetic energy increases from 0J to 196J then 0J when it hits the ground. A gain of 196J of EK is balanced by the loss of 196J of GPE,
This is true..

hence the total/net work done is zero.
.. but this is false. In order for the nett work to be 0, then *another* force would have to be acting. You can't count the loss of GPE as somehow doing work against gravity.

Consider a force F of 10 N pulling a block across a rough surface. As the block moves, it feels a frictional force G of -10 N (i.e. in the opposite direction to F). When it has moved 1 m, F has done work of 10 J on the block, and G has done work of -10 J on the block. The nett work done on the block is 0 J in this scenario. We can see that from the fact that block's KE has not increased.

Of course, we have conservation of energy here, too. In this case, the work done by F has ended up heating up the ground though, by 10 J's worth of energy.

In the original case, all of the work done by gravity ends up in the KE of the mass. That's just what the work-energy theorem states, of course.

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Updated: April 18, 2016
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