Coefficient of kinetic friction and constant acceleration

[addy899]

Homework Statement

A child's slide is flat, 5 meters long, and sloped down so that the top is 3 meters higher than the bottom. A child slides down, starting from rest, and has v = 6 m/s at the bottom. Find the coefficient of kinetic friction, μ_k is:

Homework Equations

F=ma
f=μ_kF_N
v²=v₀²+2aΔx

The Attempt at a Solution

use above equation to find constant a=3.6m/s²

3-4-5 triangle, so θ= 30°

free body diagram gives:
F_N=mgcos30

and using f=ma,
F_net=3.6m = mgsin30-μ_kF_N

the masses cancel out to give 3.6 = gsin30-μ_kgsin30

solving for μ_k gives 0.153

my key says the answer is 0.3

 

[haruspex]

3-4-5 triangle, so θ= 30°

Think that bit through again.

 

[ap123]

3-4-5 triangle, so θ= 30°

The angle here is not correct.
You don't even need to calculate the value for the angle since you already have the length of the sides.

the masses cancel out to give 3.6 = gsin30°-μ_kgsin30°

The second sin here is not correct - it should be cos as you worked out previously.

 

[addy899]

θ=36.9°

ok, since we already know the length of the sides, we get

3.6 = 5 - 5cos(36.9)μ_k

5 is the force from gravity parallel to the slide

5cosθ is the normal force perpendicular to the slide

this gives me μ_k=.35

 

[ap123]

3.6 = 5 - 5cos(36.9)μ_k

this gives me μ_k=.35

3.6 = gsin(θ)-μ_kgcos(θ)
gives
3.6 = (10)*(3/5) - μ_k*(10)*(4/5) (don't need to work out θ)
gives 0.3