In a circus performance, a monkey is strapped to a sled and both are given an in
ID: 1539997 • Letter: I
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 22.0 degree inclined track. The combined mass of monkey and sled is 22 kg, and the coefficient of kinetic friction between sled and incline is 0.20. How far up the incline do the monkey and sled move? While running, a person dissipates about 0.60 J of mechanical energy per step per kilogram of body mass. If a 66 kg person develops a power of 69 W during a race, how fast is the person running? (Assume a running step is 1.5 m long)Explanation / Answer
The frictional force acting against the motion is F
= (mu)x mg x cos a.
= 0.2 x 22 x 9.8 x cos 22 = 39. 97 N.
The net force acting along the inclined plane
= (mg sin a +39.97) Since both the forces are acting opposite to the motion.
= 22 x 9.8 x sin 22+ 39.97
= 120.6 N.
If x is the distance moved along the plane, the work done by this net force is Fx.
This work done is used to remove the kinetic energy of the body = 0.5 m v v.
Fx = 0.5 m v v
x = 0.5 x 22 x 3 * 3/ 120.6 = 0 . 82 m.
B.
Remember 1 W = 1 J/s, so if a 66 kg person develops a power of 69W during a race he's dissipating 69J/s.
Dividing by his body mass, on a per kilogram basis he's expounding (69J/s)/66 = 1.045 J/s.
Dividing by 0.60 J, the number of joules dissipated per second, he takes 1.741 steps per second, and multiplying by the distance of a step, 1.5 m, tells you that per second, he travels 2.61 m.
So his speed is 2.61 m/s.