# Thread: Work done on coin.

1. Supposing you have these two scenarios:
String is attached to a coin.
(A) You raise the coin vertically from a horizontal table through a distance of 10cm.
(B) Pull the coin up a smooth book that is tilted with respect to the table until its vertical distance changes by 10cm.

The coin moves at constant speed in both these cases.

I would like to know whether there is more work done in case B, because the coin moves a greater distance, since it is basically moving up the hypotenuse?....I understand that W=Fd, and hence the greater distance travelled. However since F=ma, and it is moving at constant speed, hence a=0, so there is no force?

When i tried this out, it felt the same to me in both cases.  2.

3. Originally Posted by johnnymeboy
Supposing you have these two scenarios:
String is attached to a coin.
(A) You raise the coin vertically from a horizontal table through a distance of 10cm.
(B) Pull the coin up a smooth book that is tilted with respect to the table until its vertical distance changes by 10cm.

The coin moves at constant speed in both these cases.

I would like to know whether there is more work done in case B, because the coin moves a greater distance, since it is basically moving up the hypotenuse?....I understand that W=Fd, and hence the greater distance travelled. However since F=ma, and it is moving at constant speed, hence a=0, so there is no force?

When i tried this out, it felt the same to me in both cases.
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Assuming that the book is smooth enough to be frictionless the work will be the same. This is one illustration of conservation of energy,

You can also show this by resolving the force necessary to slide the coin up the slide into components normal and parallel to the plane of the book and calculst the work done (only the parallel force does work) and get the same answer.  4. Sorry but i don't really understand how the conservation of energy relates to this?  5. Originally Posted by johnnymeboy
Sorry but i don't really understand how the conservation of energy relates to this?
The work done is precisely equal to the increase in potential energy.  6. Originally Posted by johnnymeboy
However since F=ma, and it is moving at constant speed, hence a=0, so there is no force?
There is a force (i.e., the weight of the coin) even with a constant speed. The acceleration is g, the acceleration due to gravity.  7. The d in your equation for work stands for the vertical distance. If distance were 3 miles across flat wet ice, little d could still be zero.  8. Originally Posted by dalemiller
The d in your equation for work stands for the vertical distance. If distance were 3 miles across flat wet ice, little d could still be zero.
No it doesn't.

It stands for the distance over which the force is applied and in fact the equation should be writen as a vector equation and the product a dot product.

The vertical distance would be appropriate in the frictionless case, where gravity is the only resisting force, but hot otherwise.

Better yet, it is the line integral of the force over the trajectory.  9. there is force and force exists because of acceleration due to gravity and also because of the fact that the coin has to start from rest from 0 to a particular value.
if force doesnt exist then the coin cannot move  10. Originally Posted by dalemiller
The d in your equation for work stands for the vertical distance. If distance were 3 miles across flat wet ice, little d could still be zero.
Sorry folks, the question did not fit my answer. Cheating, I knew by heart that in these latitudes, work is pegged to increased elevation however it was acquired.  Bookmarks
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