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?

1 Answer
Jun 14, 2016

Answer:

#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]