Question

a 3 - kg block sits on a horizontal surface; a 7 - kg block is subsequently tied to the 3 - kg block using a massless cable as shown below. note that the 7 - kg block sits on a 30° incline and that all of the contact surfaces are “rough” as indicated by the coefficients of friction listed in the diagram. when the system is released from rest, the system will begin to move (take this as a given).

• determine the acceleration of each block (20pts)
• determine how fast the 7 - kg block will be moving 3 seconds after being released (5pts)
• determine the tension in the cable joining the two blocks together (5pts)

μs = 0.10/μk = 0.20

Explanation:

Step1: Analyze forces on each block

For the 3 - kg block on the horizontal surface, the frictional force $f_1=\mu_k N_1$. Since $N_1 = m_1g$ ($m_1 = 3$ kg and $g = 9.8$ m/s²), $f_1=\mu_k m_1g$. Let the acceleration of the system be $a$ and the tension in the cable be $T$. For the 7 - kg block on the incline, the gravitational - force component along the incline is $m_2g\sin\theta$ and the frictional force is $f_2=\mu_k N_2$, where $N_2=m_2g\cos\theta$ ($m_2 = 7$ kg and $\theta = 30^{\circ}$).

Step2: Apply Newton's second law to 3 - kg block

According to Newton's second law $F = ma$, for the 3 - kg block, $T - f_1=m_1a$. Substituting $f_1=\mu_k m_1g$, we get $T-\mu_k m_1g=m_1a$.

Step3: Apply Newton's second law to 7 - kg block

For the 7 - kg block, $m_2g\sin\theta - T - f_2=m_2a$. Substituting $f_2=\mu_k m_2g\cos\theta$, we have $m_2g\sin\theta - T-\mu_k m_2g\cos\theta=m_2a$.

Step4: Solve for acceleration $a$

Add the two equations from Step2 and Step3:
\[

$$\begin{align*} m_2g\sin\theta - \mu_k m_1g-\mu_k m_2g\cos\theta&=(m_1 + m_2)a\\ a&=\frac{m_2g\sin\theta-\mu_k m_1g-\mu_k m_2g\cos\theta}{m_1 + m_2} \end{align*}$$

\]
Substitute $m_1 = 3$ kg, $m_2 = 7$ kg, $\mu_k = 0.20$, $\theta = 30^{\circ}$, and $g = 9.8$ m/s²:
\[

$$\begin{align*} a&=\frac{7\times9.8\times\sin30^{\circ}-0.20\times3\times9.8 - 0.20\times7\times9.8\times\cos30^{\circ}}{3 + 7}\\ &=\frac{7\times9.8\times0.5-0.20\times3\times9.8-0.20\times7\times9.8\times\frac{\sqrt{3}}{2}}{10}\\ &=\frac{34.3 - 5.88-11.87}{10}\\ &=\frac{34.3-(5.88 + 11.87)}{10}\\ &=\frac{34.3 - 17.75}{10}\\ &=1.655\approx1.66\text{ m/s}^2 \end{align*}$$

\]

Step5: Find the velocity of 7 - kg block after $t = 3$ s

Using the kinematic equation $v=v_0+at$, with $v_0 = 0$ m/s, $a = 1.66$ m/s², and $t = 3$ s. Then $v=0+1.66\times3 = 4.98$ m/s.

Step6: Find the tension $T$

Substitute $a = 1.66$ m/s² into the equation for the 3 - kg block $T-\mu_k m_1g=m_1a$.
\[

$$\begin{align*} T&=m_1a+\mu_k m_1g\\ &=3\times1.66+0.20\times3\times9.8\\ &=4.98 + 5.88\\ &=10.86\text{ N} \end{align*}$$

\]

Answer:

• Acceleration of each block: $1.66$ m/s²
• Velocity of 7 - kg block after 3 s: $4.98$ m/s
• Tension in the cable: $10.86$ N