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

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?

in progress 0
Ngọc Hoa 1 week 2021-07-22T08:10:25+00:00 1 Answers 1 views 0

Answers ( )

    0
    2021-07-22T08:11:37+00:00

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )