Tired of being chased by a jaguar, you set a trap. Hoping to drop it on the jaguar, you try to push a

44.0 kg stone boulder off of the edge of a cliff that slopes down at an angle of 15.0°. Being weak with

hunger, the best you can do is push the boulder with a force of 222 N. The coefficient of kinetic friction

between the boulder and the ground is is 0.700. (Ignore static friction.)

What is the acceleration of the boulder while you push it down the incline?

Answer: acceleration = 3.27m/s^2

Explanation:

Given that the

Mass M = 44kg

Angle Ø = 15 degree

Coefficient of friction ų = 0.7

Force F = 222N

F – Fr = ma …… (1)

Where Fr = frictional force

Fr = ųN

N = normal reaction = mg

Fr = ųmgsinØ

Fr = 0.7 × 44 × 9.81 × sin 15

Fr = 78.2N

Substitutes Fr, F and M into equation one.

222 – 78.2 = 44a

143.79 = 44a

Make a the subject of formula

a = 143.79/44

Acceleration a = 3.27 m/s^2