# Answer to Question #16853 in Mechanics | Relativity for gg

Question #16853

A 206-kg crate is pushed horizontally with a force of 705 N. If the coefficient of friction is 0.20, calculate the acceleration of the crate.

Expert's answer

The friction force acting on a crate is

F(friction) = μ*m*g = 0.20*206[kg]*9.81[m/s²] = 404.17 [N].

The net force acting on a crate is

F(net) = F(pushing) - F(friction) = 705[N] - 404.17[N] = 300.83 [N].

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

a = F(net)/m = 300.83[N]/206[kg] = 1.46 [m/s²].

F(friction) = μ*m*g = 0.20*206[kg]*9.81[m/s²] = 404.17 [N].

The net force acting on a crate is

F(net) = F(pushing) - F(friction) = 705[N] - 404.17[N] = 300.83 [N].

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

a = F(net)/m = 300.83[N]/206[kg] = 1.46 [m/s²].

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