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

Answer on Question #72536, Physics / Mechanics | Relativity

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

Solution:

Free body diagram for body

Newton Second Law:

$$\sum \vec {F} _ {i} = m \vec {g} + \vec {N} + \vec {F} + \vec {f _ {k}} = m \vec {a}$$

The friction force acting on the body:

$$f _ {k} = \mu_ {k} N = \mu_ {k} m g \cos \alpha$$ $$m a = F - m g \sin \alpha - \mu_ {k} m g \cos \alpha$$ $$a = \frac {F - m g \sin \alpha - \mu_ {k} m g \cos \alpha}{m}$$ $$a = \frac {1 8 0 N - 3 0 k g \times 9 . 8 1 \frac {m}{s ^ {2}} \times \sin 3 0 {}^ {\circ} - 0 . 2 2 5 \times 3 0 k g \times 9 . 8 1 m / s ^ {2} \cos 3 0 {}^ {\circ}}{3 0 k g} = - 0. 8 1 m / s ^ {2}$$

This mean that crate will stay at rest. In other words, we can see, that force $F = 180N$ compensates weigh projection on x-axis $mg\sin \alpha \approx 150N$ , but it is not enough to overcome the friction force $f_{kmax} = \mu_{k}mg\cos \alpha \approx 58.45N$ .

Answer: crate stays at rest

Answer provided by https://www.AssignmentExpert.com