# Inclined plane problem

1. Oct 6, 2016

### kubaanglin

1. The problem statement, all variables and given/known data
A boy drags his $60.0N$ sled at constant speed up a $15.0°$ hill. He does so by pulling with a $25.0N$ force on a rope attached to the sled. If the rope is inclined at $35.0°$ to the horizontal,

(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. Relevant equations
See below

3. The attempt at a solution
a)
Mass of sled: $\frac {60N}{9.8m/s^2} = 6.12kg$
Force of gravity on sled parallel to inclined plane: $mg\sin15° = 15.52N$
Normal force on sled: $mg\cos15° = 57.93N$
Force of boy on sled: $25N\cos20° = 23.49N$
I assumed that the boy was pulling the sled $35°$ from the ground, so $20°$ from the inclined plane.
Force of friction: $23.49N-15.53N = 7.97N$
$μ = \frac {7.97N}{57.93N} = 0.14$

b)
The mass of the boy is not given. Can I still figure out the acceleration?

2. Oct 6, 2016

### TSny

Your calculation of the friction force looks correct. However, your expression for the normal force is not correct. Be sure to draw a good free body diagram.
Yes.

3. Oct 6, 2016

### kubaanglin

Mass of sled: $\frac {60N}{9.8m/s^2} = 6.12kg$
Force of gravity on sled parallel to inclined plane: $mg\sin15° = 15.52N$
Normal force on sled: $mg\cos15° - 25N\cos20° = 57.93N - 23.49N = 34.44N$
Force of boy on sled: $25N\cos20° = 23.49N$
Force of friction: $23.49N-15.53N = 7.97N$
$μ = \frac {7.97N}{34.44N} = 0.23$

I believe I fixed the normal force.

I am a little confused on how to incorporate the friction coefficient without needing to know the mass.

4. Oct 6, 2016

### TSny

Almost. Check to see if you're using the correct trig functions.

Set up the equations with symbols and see what happens.