Question

A box with an initial speed of `4 m/s` is moving up a ramp. The ramp has a kinetic friction coefficient of `1/3` and an incline of `(2 pi )/3`. How far along the ramp will the box go?

Answer 1

`Delta x = (4 sqrt[3])/g`[m]

Explanation:

The initial speed is `v_0` so the initial kinetic energy is given by

`E_k = 1/2mv_0^2`

The box friction work loses along the ramp are given by

`mu xx mg xx sin(theta) Delta x`

with `mu` the kinetic friction coefficient, `theta` the ramp incline, and `Delta x` the covered distance along the ramp.

So we can equate

`1/2mv_0^2 = mu xx mg xx sin(theta) Delta x + m g sin(theta) Delta x`

Solving for `Delta x`

`Delta x =(1/2 v_0^2)/((mu+1) sin(theta)g ) = (4 sqrt[3])/g`[m]