Conservation of Energy Problem - How far does each can slide?

In a hardware store, paint cans, which weigh 46.0 N
each, are transported from storage to the back of the
paint department by placing them on a ramp that is
inclined at an angle of 24.0 deg above the horizontal. The
cans slide down the ramp at a constant speed of 3.40
m/s and then slide onto a table made of the same
material as the ramp. How far does each can slide on
the table’s horizontal surface before coming to rest?

2 Answers
Jul 7, 2016

#d= 12.9821/g#

Explanation:

The paint cans glide without acceleration so

#mu m g cos(alpha)=m g sin(alpha)#

where

#m# can mass
#g# gravity acceleration
#mu# kinetic friction coefficient
#alpha# ramp angle

from this relationship we get

#mu = tan(alpha) = 0.445229 #

When in the horizontal table the movement has initial velocity #v_0#
so the can carries a kinetic energy of

#1/2mv_0^2# which is lost against the friction losses

#1/2mv_0^2 = mu m g d# where #d# is the glided distance over the table until repose.

so

#d = 1/2 v_0^2/(mu g) = 12.9821/g#

Jul 10, 2016

#1.32m#, rounded to two decimal places.

Explanation:

The paint cans slide down with constant speed. This means there is no acceleration and upwards force due to friction is equal and opposite to the #sin theta # component of gravitational force.
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If #mu# is the coefficient of kinetic friction and other quantities as shown in the figure
#f=mu N=mu m g costheta#

#=>mu m g costheta=m g sintheta#
We obtain
#mu = tan 24^@ = 0.44523 #, rounded to five decimal places

While moving on the horizontal table the can has initial velocity #u#. The KE due to this speed is lost in doing work against the force of friction.

If #s# is the distance moved on the table before coming to rest, work done against the force of friction
#=(mu m g)cdot s#
(material of ramp and table being same, has same coefficient of kinetic friction.)
Equating it with the KE we obtain
#(mu m g)cdot s=1/2m u^2#
#=> s=(1/2m u^2)/(mu m g)#
#=> s= u^2/(2mu g)#
Inserting given, calculated quantities and taking #g=9.81ms^-2# we get
#s= (3.40)^2/(2xx0.44523 xx9.81)#
#=>s=1.32m#, rounded to two decimal places.