# Calculate the total work done on the sled.

1. Jan 12, 2010

### noman.21

1. The problem statement, all variables and given/known data

Jason walks 50 meters at constant velocity(towards the right) along a 1000 meter long hard an flat snowy surface with Uk = 0.30 while pulling on a rope attached to an empty sled of mass 10 kg. The sled is observed to follow close behind Jason. The rope remains horizontal(parallel to the surface) and taught the entire time. Just after Jason has finished walking 50 meters, he drops the rope out of hi hand and continues to run another 75 meters at the same constant velocity. The sled moves another 40 meters before coming to a full stop. Calculate the total work done on the sled during the time Jason and the sled move 50 meters together.

2. Relevant equations
Fn = ma
Fk = Uk x Fn
W=F Cos @ D
Wf = Wa + Wfk

3. The attempt at a solution

I have an attempt at it but its been marked aqwardly so I don't know which parts are right.

Last edited: Jan 12, 2010
2. Jan 12, 2010

### rl.bhat

Hi noman.21, welcome to PF.
Show your calculations to know which part of your calculation is wrong.

3. Jan 12, 2010

### noman.21

Hi, Okay well. I sort of didn't know how to attempt the problem, what I wrote above was an accident I was reading another question when I wrote that, but I was just curious about this problem. IS there any way you could help? Its not exactly homework, It was a question on a test I didn't answer and I have an exam tomorrow, just wanted to be clear on the process.

Last edited: Jan 12, 2010
4. Jan 12, 2010

### rl.bhat

First of all find the frictional force. Once the sled start moving, it will move with uniform velocity when the applied force is equal to the frictional force.

5. Jan 12, 2010

### noman.21

the firctional force is Fn x Uk = Fk

so (10kg x 9.8) (0.30) = Fk
= 29.4

But if the applied force is equal to the frictional force, wont it be stationary?

6. Jan 12, 2010

### rl.bhat

When the work is done on a body, its kinetic energy or potential energy must increase. But in the problem the sled is moving with constant velocity up to 50 m. So the net work done during this motion is ........?

7. Jan 12, 2010

### noman.21

The work done during the motion, would it be the Wa + Wf

Wa + (29.4 x cos 180 x 50m)?

8. Jan 12, 2010

### rl.bhat

No. Net work done is zero.