Time to Stop Sliding

## Question:

A 5 kg block is released from rest on a smooth 30 degree ramp at
a vertical height of 3m above a rough floor. The coefficient of
kinetic friction of the floor and crate is 0.2. Determine the
total time from the release till the block stops sliding. Also
find the distance the block slides along the floor.
## Answer:

Based on the conservation of energy, the kinetic energy of the
block at the foot of the ramp, assuming a smooth transition to
horizontal motion, is the m*g*h where m is 3 kg, g is 9.8
m/s^{2} and h is 3 meters. This gives us initial kinetic
energy of 147 Joules. The block will slide until this energy is
used up by the work of friction. The work of friction is the
force of friction times the distance traveled. The force of
friction is the normal force, m*g, times the coefficient of
friction, 0.2. This gives us the equation,
147=5*9.8*0.2*d where d is the distance slid along the
floor. Solving for d I get 15 meters.
The acceleration of the block while on the floor is only due
to the force of friction so it is constant at
5*9.8*.2/5 = 1.96m/s^{2}. Under constant
acceleration the distance traveled
d=1/2*a*t^{2}, so
15=.98*t^{2},
t^{2}=15.306, t=3.91s for
the slide along the floor. The time to slide down the ramp is
given by the same relationship where d is now the length of the
ramp and a is the component of the acceleration of gravity along
the ramp. The length of the ramp is the height divided by the
sine of the ramp angle or 6 meters. The acceleration along the
ramp is 9.8 times the sine of the ramp angle or
4.9m/s^{2}. So
6=1/2*4.9*t^{2},
t^{2}=2.45, t=1.56s. Total
time is 5.47s.

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