In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 2.0 m/s up a 20.0° inclined track. The combined mass of monkey and sled is 24 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

2.65 m

Explanation:

From work-kinetic energy principle,

workdone by friction + workdone by gravity on sled = kinetic energy change of sled

Let h be the  vertical height moved and d the distance moved along the incline. h = dsinθ   where θ is the angle of the incline = 20°.

The workdone by gravity on the sled is mghcos180 = -mgh = -mgdsinθ

The frictional force = -μmgcosθ where μ = 0.20 and the work done by friction = -μmgcosθd

The kinetic energy change = 1/2m(v₂² - v₁²) where v₁ = initial speed = 2.0 m/s and v₂ = final speed = 0 m/s (since the sled stops)

So, -mgdsinθ - μmgcosθd = 1/2m(v₂² - v₁²)

-gd(sinθ - μcosθ) = 1/2(v₂² - v₁²)

d = (v₂² - v₁²)/-2g(sinθ - μcosθ)

substituting the values for the variables,

d = (0 - 2²) /[- 2 × 9.8(sin20 - 0.2cos20)]

d = -4 /[- 2 × 9.8(sin20 - 0.2cos20)]

d = -4/ - 1.51 = 2.65 m

It has moved up the incline a distance of 2.65 m.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

read more on -