Question

A 25 kg crate is on surface with a coefficient of sliding friction of 0.2. A 90 Newton force is applied to move the crate to the right. What is the acceleration?

Answer 1

The acceleration of the crate is $1.638 \mathrm{m} / \mathrm{s}^{2}$ if a crate weighs 25 kg with a coefficient of sliding friction 0.2

Explanation:

Given that,

Mass of the crate (m) = 25kg

Coefficient of friction (µ) = 0.2

$\mathrm{F}_{\text {applied }}=90 \mathrm{N}$

Gravitational acceleration is $9.8 \mathrm{m} / \mathrm{s}^{2}$

We know that when the force is applied and frictional force acts in the body the net force is written as $\Sigma \mathrm{F}=\mathrm{F}_{\text {applied }}-\mathrm{F}_{\text {frict }}$

$\mathrm{F}_{\text {frict }}=\mu \mathrm{N} \quad(\mathrm{N}=\mathrm{mg})$

$\Sigma \mathrm{F}=\mathrm{F}_{\text {applied }}-\mu \mathrm{mg}$

According to newton second law

$\Sigma F=m a$

$a=\frac{\Sigma \mathrm{F}}{m}$

$a=\frac{F_{a p p l i e d}-\mu m g}{m}$

$a=\frac{90-(0.2 \times 25 \times 9.81)}{25}$

$a=\frac{90-49.05}{25}$

$a=\frac{40.95}{25}$

$a=1.638 \mathrm{m} / \mathrm{s}^{2}$

Therefore acceleration of crate is $1.638 \mathrm{m} / \mathrm{s}^{2}$.