TPR Ch5 #3 Practice Problem - Work done by Gravity

[sleepinballer]

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3. A young child is sliding down a hill at an incline of 30 degrees on a sled with a total combined mass of 10kg, If the coefficient of friction between the hill and the sled is 0.3 and the length of the hill is 50m, how much work has been done by gravity when the child reaches the bottom of the hill.

A. 1000J
B. 2500J
C. 3535J
D. 4330J

Finding the work done by gravity is no problem for me just use Fdsin30. My question is how come we don't take into account the "work done by friction". Does friction not do work? Friction is pointing the other direction of the the way the box is sliding. So in my mind we need to subtract the work done by friction to get the actual work done by gravity?

Can any one help please?

 

[cheesier]

It's Fdsin30 for the gravitational potential energy because gravitational force is in the y direction so you need to calculate the displacement on that same axis.

Work done by friction is not counted because it's not asking for total work done, just work done by gravity. If a question asked you to calculate the frictional force on the object you wouldn't worry about including the gravitational force somewhere in your answer. It's the same here.

 

[sleepinballer]

Thank you so much for your response!!! Why would the frictional force of the object not include the force done by gravity??? Is it because they are perpendicular to each other?? I mean the way i see is that total work by gravity is being taken away by the work done by friction...so why would they not by subtracted.

 

[cheesier]

They're two different forces, so the work done by each one is separate. If the question asked you for the work done by friction you would not include gravity anywhere in your answer; likewise for this question.

 

[sleepinballer]

Thanks alot!!