Normal force = m g cos 22
= 41 * 9.81 * cos 22 = 373 N
Force down slope = m g sin 22 = 151 N
time to slide 40 m = 40/3.5
acceleration is zero so friction force = 151 N
373 mu = 151
so
mu = 151/373
A child on a sled (total mass 41 kg) slides down a hill inclined at an angle θ = 22° at a constant speed 3.5 m/s.
(a) How long does it take the child to travel 40 meters down the slope?
(b) What is the frictional force on the sled?
(c) What is the coefficient of friction between the sled and the slope?
(d) The child reaches the bottom of the hill and slides along the horizontal muddy ground to rest. If the coefficient of friction is no different on the horizontal surface, how far does the sled go before coming to rest?
1 answer
1 answer