Question

In the arrangement shown in figure, $m_A=m_B=2kg$ String is massless and pulley is frictionless. Block B is resting on a smooth horizontal surface, while friction coefficient between blocks A and B is $\mu =0.5$. The maximum horizontal force F that can be applied so that block A does not slip over the block B is

• A. $25N$
• B. $40N$
• C. $30N$
• D. $20N$

Answer 1

Answer: (D)

Maximum value of friction on $A,f_{\text{max}}=\mu mg$

$=0.5\times 2\times 10=10N$

Hence, the block B is subjected to this frictional force in the forward direction. Since the block A does not slip on B, both of them move with a common acceleration. The acceleration of B is the same as the acceleration of A

To determine acceleration of ‘a ’, we find that $m_B{a}_{\text{max}}=f_{\text{max}}$

$a_{\text{max}}=\frac{10}{2}=5m/s^2$

$F_{\text{max}}=\left(m_A+m_B\right)a_{\text{max}}$
$=4\times 5=20N$