Calculate the total work done on the sled.

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The discussion focuses on calculating the total work done on a sled being pulled by Jason over a distance of 50 meters. The sled, with a mass of 10 kg, experiences a frictional force calculated using the coefficient of kinetic friction (Uk = 0.30). Participants clarify that while Jason pulls the sled, the net work done is zero because the sled moves at a constant velocity, indicating that the applied force equals the frictional force. The calculations involve determining the frictional force and understanding that work is done only when there is a change in kinetic or potential energy. Ultimately, the conclusion is that the net work done on the sled during this motion is zero.
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Homework Statement



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.

Homework Equations


Fn = ma
Fk = Uk x Fn
W=F Cos @ D
Wf = Wa + Wfk

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:
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Hi noman.21, welcome to PF.
Show your calculations to know which part of your calculation is wrong.
 
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:
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.
 
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, won't it be stationary?
 
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 ...?
 
The work done during the motion, would it be the Wa + Wf

Wa + (29.4 x cos 180 x 50m)?
 
No. Net work done is zero.
 

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