# Answer to Question #20621 in Mechanics | Relativity for Bobby

Question #20621

A 34kg crate is is pushed with a horizontal force of 70N at a steady pace. Determine the value of the kinetic friction and the coefficient of kinetic friction.

Expert's answer

As the crate is moving at a steady pace, it's acceleration has zero value and according to the second Newton's law we can conclude that the friction force has the same value as the pushing force and is directed oppositely. So,

F(friction) = 70[N].

Let's find the coefficient of kinetic friction μ. We know that

F(friction) = μ·F(normal),

therefore

μ = F(friction)/F(normal) = 70[N]/(34[kg]*9.8[m/s²]) ≈ 0.21.

F(friction) = 70[N].

Let's find the coefficient of kinetic friction μ. We know that

F(friction) = μ·F(normal),

therefore

μ = F(friction)/F(normal) = 70[N]/(34[kg]*9.8[m/s²]) ≈ 0.21.

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