- #1

songoku

- 2,328

- 336

- Homework Statement
- A sledge of mass 50 kg sits on a snowy surface. It is pulled horizontally for 10 m against a frictional force of 200 N, then it is pulled horizontally across ice for 10 m. There is no friction between the ice and the sledge. It is lifted up vertically by 1 m and finally carried back at a constant speed to where it started. During which stage of its journey is most work done on the sledge?

A being carried back 20 m at constant speed

B being lifted up 1 m

C being pulled 10 m across ice

D being pulled 10 m across snow

- Relevant Equations
- W = Fd

W = mgh

The answer is (D), but I don't understand why.

Option (A) is wrong because the work done = 0. Then, I divide the motion into 3 parts:

1) motion on snowy surface

Since the sledge is being pulled horizontally (let assume to the right), there will be tension force T to the right and friction force of 200 N to the left so the work done will be (T - 200) x 10

2) motion across ice

The sledge is still pulled horizontally so I assume the value of T is the same as previous motion so the work done is T x 10

3) lift up the sledge

Work done = GPE

I assume again during motion (1) the sledge is not decelerating so T is at least 200 N so the most work done is during motion (2) so my answer is (C).

I can not imagine a case where the work done in motion (1) is more than in motion (2).

Where is my mistake? Thanks

Option (A) is wrong because the work done = 0. Then, I divide the motion into 3 parts:

1) motion on snowy surface

Since the sledge is being pulled horizontally (let assume to the right), there will be tension force T to the right and friction force of 200 N to the left so the work done will be (T - 200) x 10

2) motion across ice

The sledge is still pulled horizontally so I assume the value of T is the same as previous motion so the work done is T x 10

3) lift up the sledge

Work done = GPE

I assume again during motion (1) the sledge is not decelerating so T is at least 200 N so the most work done is during motion (2) so my answer is (C).

I can not imagine a case where the work done in motion (1) is more than in motion (2).

Where is my mistake? Thanks