Question

a 5.00 kg block placed on an inclined plane angled 30.0 degree above the horizontal. a rope is attached to the mass, goes through a pulley, and attached to a 6.00 kg ball which is hanging off the edge. the plane/floor has a frictional coefficient of 0.200 a.) find the acceleration of the system. b.) what is the tension force in the rope? a m1 = 5kg m2 = 6kg

Explanation:

Step1: Analyze forces on block on inclined - plane

For the 5.00 kg block ($m_1$) on the inclined plane, the normal force $N$ is given by $N = m_1g\cos\theta$, where $g = 9.8\ m/s^2$ and $\theta=30.0^{\circ}$. So, $N = 5\times9.8\times\cos30^{\circ}=5\times9.8\times\frac{\sqrt{3}}{2}\approx42.43\ N$. The frictional force $f=\mu N$, with $\mu = 0.200$. So, $f = 0.2\times42.43 = 8.49\ N$. The component of gravitational force along the plane is $m_1g\sin\theta=5\times9.8\times\sin30^{\circ}=24.5\ N$. Let the tension in the rope be $T$ and the acceleration of the system be $a$. According to Newton's second - law, for the block on the inclined plane, $T - m_1g\sin\theta - f=m_1a$.

Step2: Analyze forces on hanging ball

For the 6.00 kg ball ($m_2$), the force equation according to Newton's second - law is $m_2g - T=m_2a$.

Step3: Solve the system of equations

We have the system of equations:
\[

$$\begin{cases} T - m_1g\sin\theta - f=m_1a\\ m_2g - T=m_2a \end{cases}$$

\]
Add the two equations together: $m_2g - m_1g\sin\theta - f=(m_1 + m_2)a$.
Substitute the values: $m_1 = 5\ kg$, $m_2 = 6\ kg$, $g = 9.8\ m/s^2$, $\theta = 30^{\circ}$, and $f = 8.49\ N$.
\[

$$\begin{align*} 6\times9.8-5\times9.8\times\sin30^{\circ}-8.49&=(5 + 6)a\\ 58.8-24.5 - 8.49&=11a\\ 25.81&=11a\\ a&=\frac{25.81}{11}\approx2.35\ m/s^2 \end{align*}$$

\]

Step4: Find the tension in the rope

Substitute $a$ into the equation $m_2g - T=m_2a$. So, $T=m_2(g - a)$.
\[

$$\begin{align*} T&=6\times(9.8 - 2.35)\\ &=6\times7.45\\ &=44.7\ N \end{align*}$$

\]
A.

Answer:

The acceleration of the system is approximately $2.35\ m/s^2$.
B.