Question

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 25.0° inclined track. The combined mass of monkey and sled is 16 kg, and the coefficient of kinetic friction between sled and incline is 0.20. How far up the incline do the monkey and sled move?

_______m

Answer 1

Initial speed of the monkey and sled = u = 3.0 m/s

final speed = o >> as both come to rest

Now,

both are moving up

mg sin 25 is acting downwards

uk[mg cos 25] is acting downwards as friction is opposite to direction of motion

So, the net force in the upward direction,

ma = - {mg sin 25} + {- uk mg cos 25}

=> a (up) = - g sin 25 - uk g cos 25

= -9.8*0.42 - 0.2*9.8*0.91

= -5.90 m/s^2

let distance traveled be (d) UP the incline

v^2 = u^2 + 2 a d

0 = 3^2 - 2 *5.90* d

d = 9/(2*5.90) = 0.76 meter

Therefore, distance traveled by the monkey and the sled in the upward direction of the incline, d = 0.76 meter (Answer)