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?

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

. west