Question

Block $A$ of mass $35kg$ is resting on a frictionless floor. Another block $B$ of mass $7kg$ is resting on it as shown in figure. The coefficient of static friction between the blocks is $0.5,$ while coefficient of kinetic friction is $\mathrm{0.4.}$ If a force of $100N$ is applied to block $B$ acceleration of block $A$ will be (Take $g=10ms{}^{-2})$

• A. $0.8m{s}^{-2}$
• B. $2.4m{s}^{-2}$
• C. $0.4m{s}^{-2}$
• D. $4.4m{s}^{-2}$

Answer 1

Answer: (A)

Here, $m_A=35kg,m_B=7kg,{\mu }_s=0.5,{\mu }_k=0.4$
$F=100N,g=10m{s}^{-2}$
Static frictional force on $B$
$f_s={\mu }_s{m}_{B}g=0.5\times 7\times 10=35N$
As $F>f_s$, therefore $A$ and $B$ will move in the same direction
i.e., of applied force, but with different accelerations. Dynamic frictional force on $B$,
$f_k={\mu }_k{m}_{B}g=0.4\times 7\times 10=28N$
This will oppose the motion of $B$ and cause the motion of $A$.
For $B$, the equation of motion is $F-f_k=m_B{a}_B$
$\therefore 100-28=7a_B\text{ or }{a}_B=\frac{72}{7}=10.3m{s}^{-2}$
For $A$, the equation of motion is $m_A{a}_A=f_k$
$\therefore 35a_A=28\text{ or }{a}_A=\frac{28}{35}=0.8m{s}^{-2}$