Question

A 500-kg crate sits on a 10-degree ramp. If friction between the ramp and the crate is 800 N, what is the acceleration of the crate?

Answer 1

By Newton's second law, the net force acting on the crate parallel to the surface is

∑ F = mg sin(10°) - 800 N = ma

where m = 500 kg is the mass of the crate and a is the acceleration.

Solve for a :

a = ((500 kg) (9.80 m/s^2) sin(10°) - 800 N) / (500 kg)

a ≈ 0.102 m/s^2