Question

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

Answer 1

Answer on Question #73196, Physics / Mechanics | Relativity

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

Given. $m = 30.0 \, kg$; $F = 180 \, N$; $\mu_k = 0.225$; $S = 15.0 \, m$; $v_0 = 0$.

Find. $v$-?

Solution.

According to the second Newton's Low, we have

$$N - mg \cos \alpha = 0 \quad \rightarrow \quad N = mg \cos \alpha$$ $$F - F_f - mg \sin \alpha = ma \quad \rightarrow \quad F - \mu_k N - mg \sin \alpha = ma \quad \rightarrow$$ $$\rightarrow F - \mu_k mg \cos \alpha - mg \sin \alpha = ma \rightarrow a = \frac{F - \mu_k mg \cos \alpha - mg \sin \alpha}{m} =$$ $$= \frac{F}{m} - (\mu_k \cos \alpha + \sin \alpha) g = \frac{180}{30} - (\sin 30{}^\circ - 0.225 \cdot \cos 30{}^\circ) \cdot 9.8 \approx 3.01 \, m/s^2$$

So

$$S = \frac{v^2 - v_0^2}{2a} \rightarrow S = \frac{v^2}{2a} \rightarrow v = \sqrt{2aS} = \sqrt{2 \cdot 3.01 \cdot 15} \approx 9.5 \, m/s$$

Answer. $v = 9.5 \, m/s$.

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