Time to Stop Sliding


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.


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/s2 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/s2. Under constant acceleration the distance traveled d=1/2*a*t2, so 15=.98*t2, t2=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/s2. So 6=1/2*4.9*t2, t2=2.45, t=1.56s. Total time is 5.47s.

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