# Answer to Question #17632 in Mechanics | Relativity for Alexis

Question #17632

A 214-kg crate is pushed horizontally with a force of 714 N. If the coefficient of friction is 0.20, calculate the acceleration of the crate.

Expert's answer

A 214-kg crate is pushed horizontally with a force of 714 N. If the coefficient of friction is 0.20, calculate the acceleration of the crate.

The friction force acting on a crate is

F = μMg = 0.20 * 214[kg] * 9.8[m/s²] = 419.44 [N].

According to the second Newton's law the acceleration of a crate is

a = F/M = (714[N] - 419.44[N])/214[kg] ≈ 1.38[m/s²].

The friction force acting on a crate is

F = μMg = 0.20 * 214[kg] * 9.8[m/s²] = 419.44 [N].

According to the second Newton's law the acceleration of a crate is

a = F/M = (714[N] - 419.44[N])/214[kg] ≈ 1.38[m/s²].

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