Question

Which of the following statements is/are correct for the figure shown? [Take $g=10m/s^2$]

• A. The magnitude of limiting frictional force on mass $A$ is $15N$.
• B. The magnitude of limiting frictional force between mass $B$ and ground is $100N$.
• C. If force F is increasing continuously, maximum acceleration of $A$ is $3m/s^2$.
• D. If force $F$ is increasing continuously, when sliding starts between block $A$ and $B$, the minimum acceleration of $B$ relative to $A$ is $2m/s^2$.

Answer 1

Answer: (A), (B), (C)

The correct options are
A The magnitude of limiting frictional force on mass $A$ is $15N$.
B The magnitude of limiting frictional force between mass $B$ and ground is $100N$.
C If force F is increasing continuously, maximum acceleration of $A$ is $3m/s^2$.

From given,
$f_{maxA}={\mu }_s{m}_{A}g=0.3\times 5\times 10=15N{f}_{maxB}={\mu }_s^{\prime }(m_A+m_B)g=0.5\times (5+15)10=100N$

Maximum acceleration of $A$ will be when two of the blocks move together and static friction helps in moving block $A$.
$\therefore a_{maxA}=f_s/m_A=\frac{0.3\times 5\times 10}{5}=3m/s^2$

So for maximum acceleration of A, the applied force $F$ is given by,
$F=(m_A+m_B)a+\mu (m_A+m_B)g=140N$
Now when $F>140N$, then sliding will start. At this point kinetic force will act on both of the blocks, and the accelerations of the blocks are:
$a_B=\frac{140-{\mu }_k{m}_{A}g-{\mu }_k^{\prime }(m_A+m_B)g}{m_A}=\frac{140-5-80}{15}=\frac{11}{3}m/s^2$
$a_A=\frac{{\mu }_k{m}_{A}g}{m_A}=\frac{0.1\times 5\times 10}{5}=1m/s^2$
$\therefore \text{relative acceleration}=\frac{11}{3}-1=2.66m/s^2$