How far up the incline does the sled move in the following case:
In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 4.7 m/s up a 26 degree inclined track. The combined mass of monkey and sled is 22 kg, and the coefficient of kinetic friction between sled and incline is 0.37. The acceleration of gravity is 9.8 m/s^2.
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In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 4.7 m/s up a 26 degree inclined track. The combined mass of monkey and sled is 22 kg.
The gravitational force with which the monkey and the sled are attracted to the Earth is 22*9.8 N. This can be divided into two components, one parallel to the inclined track and acting downwards and the other perpendicular to the track. The perpendicular component is 22*9.8*cos 26 = 193.77 This is the normal force N with which the coefficient of kinetic friction is multipled to give the frictional force that is equal to 71.69 N.
Initially the kinetic energy of the monkey and the sled is (1/2)*22*4.7^2 = 242.99 J
Let the distance that the sled moves up the incline be D. At the top of the incline, the potential energy of the sled is 22*9.8*D*sin 26. The energy spent against the force of friction is 71.69*D
As energy is conserved: 242.99 = 22*9.8*D*sin 26 + 71.69*D
=> D = 1.462 m
The monkey and the sled move up the incline a distance equal to 1.462 m
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