Calculating Work - With Friction

[vertciel]

Hello everyone,

I am having some trouble understanding why friction is not accounted for when calculating work done, as in the following two problems.

If an object is experiencing friction, wouldn't there need to be more work done to continue moving the object?

Thank you!

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1. A father is pulling his two girls in their toboggan with a force of 500 N for a distance of 22 m. Calculate the work that would be done by the father in each of the following cases.

a) The snow provides no friction.

b) One daughter drags her hand in the snow, producing a firctional force of 500 N.

My Work:

a) Work = Force x Distance

= 500 N x 22 m

= 1.1E4 J

b) Since there is 500 N of frictional force:

Work = 1000 N x 22 m

= 2.2E4 J <-- According to textbook, this answer is wrong.

 

[TMM]

"Work done by the father."

He's not doing frictional work because he isn't friction.

 

[vertciel]

Thanks for your reply.

However, his daughter is exerting friction on the entire toboggan. Wouldn't the father have to compensate for this by doing more work?

 

[TMM]

No. You can think of it this way: the effective force will be lowered, but that force will be applied for more time, since it must go the same distance. These two effects cancel, leaving you with the same work each time.

 

[Doc Al]

Just to add a bit to TMM's explanation: Since the father pulls with the same force over the same distance, the work he does is the same in each case. But the result of that work is quite different in each case. Without friction, his work goes into accelerating the toboggan--it goes faster and faster. With friction, his work goes to overcoming friction--the sled pulls along at the same rate.

 

[Dahaka14]

This problem is making the assumption that she puts her hand out once the toboggan has an initial velocity. Therefore, the replies have been correct, meaning that this friction is called a kinetic friction.

However, if this is a static friction, where the daughter puts her hand out while the toboggan is at rest...
There is no work done in the second case because the forces cancel each other out; the toboggan does not move. The force of the father's pull is equal in magnitude but opposite in direction to the frictional force, so the vector addition yields zero. Even if you assume that it will move, since this is a non-conservative system, you must do the calculation of adding up the work done by conservative forces and non-conservative forces to get the net work. If you do this, you will get

Work done by father + Work done by friction = Net Work

500N*22m + (-500N)*22m = 0 J,

further proving that no work will be done; thus no displacement.

I am not 100% positive on this, but I'm pretty sure that if it is static, it won't move. Someone please debunk me my post is false.

 

[Doc Al]

The problem states that the toboggan moves 22 m, so we know it moves. There's nothing wrong with calculating the net work on the toboggan, but that's not what the problem asked for. No net work means no change in kinetic energy--it does not mean no displacement.

 

[Dahaka14]

Thank you for pointing that out, Doc. You are absolutely right; I didn't choose my words carefully enough. The only thing is that I know it doesn't ask for net work; it was simply an illustration of how the system would stay static if it started at rest. But you are right, thanks for correcting me.