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

Answer:

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

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 $=\text{total applied force}-\text{total friction foce}$

$1860-480$

$1380\text{N}$

$\text{force}=\text{mass}\times\text{acceleration}$

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

$\text{acceleration}=\frac{1380}{230}$

putting values

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

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