Question

A crate of mass 30.0 kg is pulled by a force of 180 N up an inclined plane which makes
an angle of 30º with the horizon. The coefficient of kinetic friction between the plane
and the crate is µk = 0.225. If the crates starts from rest, calculate its speed after it has
been pulled 15.0 m. Draw the free body diagram.

Answer 1

velocity is 9.486m/s

Answer 2

The crate remains at rest when the 180 N upward force is applied to pull the crate upward. The upward pulling force failed to overcome the frictional force and hence the crate remained in its starting point only. So its speed can’t be measure at 15 m distance with this upward force.

Explanation:

This question deals with Newton’s second law, i.e.,

$F=ma$

As the crate is pulled upward in an inclined plane, the total force acting on the crate will be a sum of normal force acting on the crate, frictional force between the crate and plane and the upward pulling force. The normal force acting on the crate can be resolved in horizontal and vertical direction force.

So it can be written mathematically as,

$\sum F=F-m g \sin \theta-\mu_{k} m g \cos \theta=m a$

The negative signs are used for normal and frictional force because it is opposite direction to the upward pulling force.

$a=\frac{F-m g \sin \theta-\mu_{k} m g \cos \theta}{m}$

So, by substituting the given values in the above equation, we get,

$a=\frac{180-(30 \times 9.8 \times \sin 30)-(0.225 \times 30 \times 9.8 \times \cos 30)}{30}$

$a=\frac{180-147-57.28}{30}$

$a=-0.81 \ \mathrm{m} / \mathrm{s}^{2}$

This indicates that the crate remains at rest when the 180 N upward force is applied to pull the crate upward. The upward pulling force failed to overcome the frictional force and hence the crate remained in its starting point only.