Question

is the maximum force $F$ that can be applied on block $m_1$, so that both $m_1\&m_2$ will move together? There is no friction between $m_1$ and the horizontal table. The coefficientof friction between ' $m_1$ ' and ' $m_2$ ' is $\mu$.

• A. $\mu m_{2}g$
• B. $\mu \left(m_1+m_2\right)g$
• C. $\mu \frac{m_1{m}_2}{\left(m_1+m_2\right)}g$
• D. $\mu m_{1}g$

Answer 1

Answer: (B)

Given, force $F$ is the applied force on mass $m_1$.

As we know that,
$F=ma$
$a=\frac{F}{m}$
where, $a$ is acceleration of system.
$m=m_1+m_2\text{ i.e. mass of system }$
$\therefore a=\frac{F}{m_1+m_2}$
and since, there should be no relative motion between $m_1$ and $m_2$
$\therefore f=\mu N_2=m_{2}a$
where, $f=$ friction force
and $N_2=$ normal reaction of mass $m_2$.
$\therefore \mu m_{2}g=m_2\frac{F}{m_1+m_2}$
$\Rightarrow F=\mu \left(m_1+m_2\right)g$