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