Solve Physics F=ma Help Problem

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Physics F=ma Help!

Homework Statement



Ok so I'm supposed to present the solution of a problem to my AP Physics class tomorrow. The thing is, I don't understand the question at all! Here is the problem (easy) and the possible answer choices. How would I go about doing this?

A 150-N sled is pulled up a 28 degree slope at a constant speed by a force of 100N parallel to the hill. What force directed up the hill will allow the sled to move downhill at a constant speed?

A) 181N
B) 170N
C) 130N
D) 141N

Homework Equations



F=ma
F=uN (Not sure if I need)
F(parallel) = W*sin(Theta)
F(perpendicular) = W*cos(Theta)

The Attempt at a Solution



Since the sled is moving up at a constant velocity, the summation of the forces would be 0. Thus the summation of the forces would be 100N - 150Nsin28 = 0. However, 100N - 150sin28 only yields 29.6N so I am assuming there is a frictional force of 29.6N. Thus I found the coefficient of friction to be 0.224 (but is that even relevant?).


What I do not get is what the problem is asking. Can someone please clarify what it is specifically asking and what I would do to get that answer? Also, if you would be kind enough, can you post the answer so I have a target answer that I can work towards. (Don't worry about me cheating because my teacher requires me to show all work or she won't even bother to look at the question.)

Thanks for all your help!
 
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You need to find with what force you need to pull up-hill so that the sled moves at a constant speed down-hill, remembering the frictional force will be in the opposite direction to the motion.
 
qspeechc said:
You need to find with what force you need to pull up-hill so that the sled moves at a constant speed down-hill, remembering the frictional force will be in the opposite direction to the motion.

Yeah but to make it go down-hill at a constant speed, you would need a net force of 0. And if u account for friction and the parallel component of weight, the answer would just be 100N and that's not one of the answer choices.