Question

A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1860 N to the right. What is the acceleration of the crate? (Hint: remember to find net force first)

Answer 1

The acceleration would be 6m/sThis is because of the formula, "f/m=a" to find the acceleration; We would need to subtract the force of the friction which equals 1380, then divide that by the mass (which was 230) to get the answer 6

Answer 2

`6\text{m}/\text{s}^2`

Step-by-step explanation:

Given: A
`230`
`\text{kg}` steel crate is being pushed along a cement floor. The force of friction is
`480`
`\text{N}` to the left and the applied force is
`1860`
`\text{N}` to the right.

To Find: What is the acceleration of the crate.

Solution:

Mass of steel crate
`=230\text{N}`

Force of friction on steel crate
`=480\text{N}`

Force applied on steel crate
`=1860\text{N}`

Now,

As force of friction is in opposite direction of force applied

Net force applied on steel crate
`=\text{total applied force}-\text{total friction foce}`

`1860-480`

`1380\text{N}`

`\text{force}=\text{mass}*\text{acceleration}`

`\text{acceleration}=\frac{\text{force}}{\text{mass}}`

`\text{acceleration}=(1380)/(230)`

putting values

`\text{a}=6\text{m}/\text{s}^2`

Hence acceleration of crate is
`6\text{m}/\text{s}^2`