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

Task #81292

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 Low is:

$$\overrightarrow {m g} + \overrightarrow {N} + \overrightarrow {F} + \overrightarrow {f _ {k}} = \overrightarrow {m a}$$ $$f _ {k} = \mu_ {k} * N = \mu_ {k} * m g * c o s \alpha (N = m g)$$ $$m a = F - m g ^ {*} \sin \alpha - \mu_ {k} ^ {*} m g ^ {*} c o s \alpha$$ $$a = (F - m g ^ {*} \sin \alpha - \mu_ {k} ^ {*} m g ^ {*} \cos \alpha) / m$$ $$a = (1 8 0 - 3 0 * 9. 8 * 0. 5 - 0. 2 2 5 * 3 0 * 9. 8 * 0, 8 7) / 3 0 = (1 8 0 - 1 4 7 - 5 7. 5 5 0 5) / 3 0 = - 0. 8 1 m / s ^ {2}$$

This mean that crate will stay at rest. Force $\approx 180\mathrm{N}$ and it compensates weigh projection on x-axis $\mathrm{mg}^{*}\sin \alpha \approx 150\mathrm{N}$ , but there is not so much to overcome the friction force, which is $\mu_{\mathrm{k}}^{*}\mathrm{mg}^{*}\cos \alpha \approx 58.45\mathrm{N}$

Answer:

Crate will stay at rest.

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