## 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?

Question

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## Answers ( )

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

∑

F=mgsin(10°) – 800 N =mawhere

m= 500 kg is the mass of the crate andais 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