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

Step-by-step 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=(F_(a p p l i e d)-\mu m g)/(m)`

`a=(90-(0.2 * 25 * 9.81))/(25)`

`a=(90-49.05)/(25)`

`a=(40.95)/(25)`

`a=1.638 \mathrm{m} / \mathrm{s}^(2)`

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