How Do You Calculate the Coefficient of Kinetic Friction on a Sled?

[NewsboysGurl91]

1. A boy drags his 60.0N sled at constant speed up a 15 degree hill. He does so by pulling with a 25.0-N force on a rope attached to the sled.If the rope is inclined at 35.0 degrees to the horizantal, (a) what is the coefficient of kinetic friction between sled and snow? (b) At the top of the hill, he jumps on the sled and slides down the hill. What is the magnitude of his acceleration down the slope?



2. uk = fk/ fn, F=ma, Fkinetic = ukN (ukmg)

http://files.tagworld.com/1b8d4ad5e8f8420aa647e517a48eef7e.jpeg That is my free body diagram.

3. Okay so I said 60N = mg cos 15 so that I would get m. Plugging in 10 for g (because of gravity), I found that the mass of the block was 6.211657082. I have no idea if that's right or not.

But I went on saying that Fn = mgcos15 - Ncos 15. I plugged in 6.21 for m, 9.8 for g, and 60 for N (Is that correct?). I got .84445 N.

Then to find fk, I found out it was equal to mgsin15 - Tsin15. Again, I plugged in the numebrs for m, g, and N, the same as before. I got that fk is .22629698076 N.

To get the coefficient of kinetic friction, I divided fk/fn, using the values I got. However, my answer was .2 something. The book says the answer is supposed to be .161. I totally have no clue as to this problem so that I would get the correct answer. Can someone help me?

 

[cristo]

The force of 60N that you have on your diagram should not be there. 60N is the weight of the sled, and so that should be in place of "mg" in your free body diagram. Then, the parallel and perpendicular components of the weight are 60sin15 and 60cos15, respectively. Try adjusting this and see what you get.

 

[NewsboysGurl91]

So what you're saying to do is Fn= 60cos15 - 60cos15 ? Then the answer is automatically 0 or undefined.

 

[cristo]

So what you're saying to do is Fn= 60cos15 - 60cos15 ? Then the answer is automatically 0 or undefined.

No.. you have another force in the free body diagram, namely the pulling force of 25N. So, your normal contact force will have contributions from both the weight of the sled and the pulling force.