Question

A 29-kg rock approaches the foot of a hill with a speed of 16 m/s. The hill slopes upward at a constant angle of 41° above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.78 and 0.2, respectively. Find the maximum height above the foot of the hill reached by the rock.

Answer 1

10.6 m

Step-by-step explanation:

Draw a free body diagram. There are 3 forces on the rock:

Normal force N pushing up normal to the surface,

Weight force mg pulling straight down,

and friction force Nμ pushing down parallel to the surface.

Sum of forces in the normal direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of forces in the parallel direction:

∑F = ma

-mg sin θ − Nμ = ma

Substitute:

-mg sin θ − mg cos θ μ = ma

-g (sin θ + μ cos θ) = a

Given initial velocity u, final velocity 0, and acceleration a, find the displacement:

v² = u² + 2as

0 = u² − 2g (sin θ + μ cos θ) s

s = u² / [2g (sin θ + μ cos θ)]

Plug in values:

s = (16 m/s)² / [2 (9.8 m/s²) (sin 41° + 0.2 cos 41°)]

s = 16.2 m

The rock slides 16.2 meters up the incline. The vertical displacement is:

h = s sin θ

h = (16.2 m) (sin 41°)

h = 10.6 m

Notice the kinetic coefficient of friction was used because the rock was in motion. Also notice that the mass of the rock ultimately did not matter.