Question

A crate is given an initial push on a horizontal floor and is imparted with an initial velocity of 5.0 m/s. Because of kinetic friction, the crate slows down and comes to stop with constant deceleration over a time period of 1.1 seconds. The mass of the crate is 5.0kg. what is the magnitude of the friction force that causes the crate to come to a stop? what is the coefficient of kinetic friction? (Assume that friction is the only force slowing the crate)

Answer 1

PART 1:

Given that:

Initial speed, V = 5.0 m/s

Final speed, u = 0 m/s

Time taken to stop, t = 1.1 s

Therefore, deceleration (magnitude) of the crate:

$\small a=\frac{v-u}{t}=\frac{5\;m/s-0\;m/s}{1.1\;s}$

$\small \therefore a=4.545\;m/s^{2}$

Mass of crate, m = 5kg

Therefore Magnitude of friction force is given by:

F = ma = (5 kg) * (4.545 m/s2)

$\small \boldsymbol{\therefore F=22.73\;N}$

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PART 2:

Friction force is given by:

$\small F=\mu_{k}mg=22.73\;N$

Therefore, Coefficient of kinetic friction is given by:

$\small \mu_{k}=\frac{F}{mg}=\frac{22.73\;N}{5\;kg\times 9.8\;m/s^{2}}$

$\small \boldsymbol{\therefore \mu_{k}=0.464}$

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