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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#339914 on Dec 2023
#339914 on Dec 2023
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