Question

A box is given a push up a 16.1° incline. When it reaches the bottom again it is only going 0.59 its original speed. Find the coefficient of kinetic friction.

*I know the answer is .06 but I don't know how!

Answer 1

It's an interesting problem.
Do you happen to be given the initial velocity as 0.7 m/s ?
I have an impression that μ is dependent on the initial velocity.

Answer 2

Unfortunately I wasn't given any other information. I'm thinking that you may have to solve in terms of variables and then the variables may cancel out?

Answer 3

Yes, I did it differently, and this time I can eliminate all other variables.
I get 0.06 m/s only if the final speed has been <i>reduced</i> by 0.59 of the original speed. If the final speed is 0.59 of the original speed, then μ=0.1396.

Let:
t=theta=16.1°
g=acceleration due to gravity
h=maximum height reached
μ=coefficient of kinetic friction
v0=initial speed
v1=return speed=0.59v0

Assume gravitational potential energy is zero at the starting (and ending) position.

Assume static friction equals kinetic friction so that box does not get "stuck" at the highest point.

Next we will determine total energy and work done at three instants,
A) at the start,
KE=(1/2)v0^2
PE=0
Work=0
Total energy,
E1 = (1/2)v0²

B) at the highest point
KE=0
PE=mgh
Work=energy dissipated=μmgcos(t)h/sin(t)
Total energy
E2 = mgh+μmgcos(t)h/sin(t)

C) at the end point,
KE=(1/2)v1²
PE=0
Work=energy dissipated=2μmgcos(t)h/sin(t)
Total energy
E3 = (1/2)v1² + 2μmgcos(t)h/sin(t)

Assuming no other energy lost/gained, then
E1=E2
E1=E3=2*mu*m*g*cos(t)*h/sin(t)/(1-0.59^2)
=>
E2=E3 ..............(1)
Solve for μ
m,g,h will all cancel out, giving
μ=0.1396 m/s

Note: if v1=(1-0.59)V0, then μ=0.060823...