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3 years ago

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A horizontal force of 93.7 N is applied to a 42.5 kg crate on a rough, level surface. If the crate accelerates at 1.03 m/s2, wha

t is the magnitude of the force of kinetic friction (in N) acting on the crate

1 answer:

vovangra [49]3 years ago

8 0

Answer:

49.925N

Explanation:

According to newton's second law of motion:

$\sum Fx = ma_x$

$\sum Fx$ is the sum of force along the x component

m is the mass of the crate

ax is the acceleration

$Fm - Fk = ma_x$

Fk is the magnitude of the force of kinetic friction

Given

Fm = 93.7

m = 42.5kg

a = 1.03m/s²

Substitute into the formula:

$93.7 - Fk = (42.5)(1.03)\\93.7-Fk = 43.775\\Fk = 93.7 - 43.775\\Fk = 49.925N$

Hence the magnitude of the force of kinetic friction (in N) acting on the crate is 49.925N

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