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
Cesareo R.
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^@)#

Related questions
Impact of this question
915 views around the world
You can reuse this answer
Creative Commons License