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# Work Done A level Question watch

1. Calculate the work done by the crane when the object of 500N is moved from A to B.
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2. The only work it has to do in this situation is against gravity so I'd say its work done is equal to its change in potential energy

Ep = mgh

but we already know the weight of the object so just multiply it by the change in height ( 500 x 40 )
3. Oh and the 500 N should be acting down and not up and is usually drawn connected to the object
4. W = distance travelled x force applied parallel to the direction of travel. What is the component of the force that acts along the line AB?

(Original post by xEPicNiceX)
Calculate the work done by the crane when the object of 500N is moved from A to B.
Not supposed to give full solutions. AFAIK we can check answers or prod you in the right direction
5. (Original post by an_atheist)
W = distance travelled x force applied parallel to the direction of travel. What is the component of the force that acts along the line AB?

Not supposed to give full solutions. AFAIK we can check answers or prod you in the right direction
So I did - Work = Force x Distance = 500N X 40m =20000Nm
I am not sure but I just assumed that there is no horizontal work done in this case?
6. (Original post by MRWAFFLE)
So I did - Work = Force x Distance = 500N X 40m =20000Nm
I am not sure but I just assumed that there is no horizontal work done in this case?
AB isnt vertical, so I dont think you can do that. I think its 500*50*0.6
7. (Original post by MRWAFFLE)
So I did - Work = Force x Distance = 500N X 40m =20000Nm
I am not sure but I just assumed that there is no horizontal work done in this case?
You are correct.

No work is done horizontally, because the vertical force has no horizontal component.
8. (Original post by an_atheist)
AB isnt vertical, so I dont think you can do that. I think its 500*50*0.6
This is not correct. The work done to a system is equal to the energy transferred, so:
work done = change in potential energy.

Your method, resolving the force along the path and multiplying by path length for WD, is also valid, but you've used the wrong trig function. Note the alternate angles (or Z-angles) formed by the geometry of the problem and try again.

You will find this approach gives the same answer as the OP, but is needlessly complicated.

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