Question

problem 8: in this problem you will do problem 6 again but the table will exert a friction force. suppose that the coefficient of friction between the table and the block is $mu_{k}=0.2$. find the acceleration of the two blocks ($a =?$) and find the tension of the string ($t =?$).

Explanation:

Step1: Set up force - equations for each block

Let the mass of the hanging block be $m_1$ and the mass of the block on the table be $m_2$. For the hanging block, using Newton's second - law $F = ma$, we have $m_1g−T = m_1a$. For the block on the table, the frictional force $f=\mu_kN$ and $N = m_2g$, and $T - f=m_2a$, so $T-\mu_km_2g=m_2a$.

Step2: Solve the system of equations

From $m_1g−T = m_1a$, we can express $T=m_1g - m_1a$. Substitute $T$ into the second equation: $m_1g - m_1a-\mu_km_2g=m_2a$. Rearrange the terms to solve for $a$:
\[

$$\begin{align*} m_1g-\mu_km_2g&=(m_1 + m_2)a\\ a&=\frac{m_1g-\mu_km_2g}{m_1 + m_2}=\frac{g(m_1-\mu_km_2)}{m_1 + m_2} \end{align*}$$

\]

Step3: Find the tension $T$

Substitute the value of $a$ into $T=m_1g - m_1a$:
\[

$$\begin{align*} T&=m_1g-m_1\times\frac{g(m_1-\mu_km_2)}{m_1 + m_2}\\ &=\frac{m_1g(m_1 + m_2)-m_1g(m_1-\mu_km_2)}{m_1 + m_2}\\ &=\frac{m_1gm_2(1 + \mu_k)}{m_1 + m_2} \end{align*}$$

\]

Answer:

Acceleration $a=\frac{g(m_1-\mu_km_2)}{m_1 + m_2}$, Tension $T = \frac{m_1gm_2(1+\mu_k)}{m_1 + m_2}$ (where $g$ is the acceleration due to gravity, $m_1$ is the mass of the hanging block, $m_2$ is the mass of the block on the table, and $\mu_k = 0.2$)