Cans of 46.0 N slide down the ramp 24.0 deg above the horizontal at a constant speed of 3.40 m/s and then slide onto a table made of the same material. How far does each slide on the table’s horizontal surface before coming to rest?

1 Answer
Dec 20, 2016

1/2 cdot 3.4^2/9.81 cdot cot(24^@)

Explanation:

Calling

theta=24^@ ramp declivity.
v = 3.40 speed along ramp.
m = 46.0 can mass.
g = 9.81 gravity acceleration.
mu kinetic friction coefficient.
d sliding distance.

When sliding with constant speed the dynamic behavior is described by the equation

-mgsin(theta)+mu N = 0

along the ramp surface, where N = m g cos(theta) is the normal force.

After resuming the ramp, the kinetic energy is given by

K=1/2m v^2

This energy is wasted against friction so

K = mu m g d which is the work produced by the friction force.

Solving for mu we get

mu = tan(theta)

Solving now for d

d = 1/2v^2/(mug) = 1/2v^2/g cot(theta) = 1/2 cdot (3.4)^2/9.81 cdot cot(24^@)