Question

Two bodies $m_1$ and $m_2$ are kept on a rough horizontal table having coefficient of friction $\mu$ and are joined by a spring. Initially, the spring is in its relaxed state. Find the minimum force $F$ which will make the other block $m_2$ move. ($k$ is the spring constant).

• A. $\mu g[m_1+\frac{m_2}{2}]$
• B. $\mu g[m_2+\frac{m_1}{2}]$
• C. $g[m_1+\frac{\mu m_2}{2}]$
• D. $g[m_2+\frac{\mu m_1}{2}]$

Answer 1

The correct option is A $\mu g[m_1+\frac{m_2}{2}]$

Motion of $m_2$ starts when elongation in the spring is $x$,

FBD, of both the blocks as shown in figure.


For the minimum force,

$kx=\mu m_{2}g$

where $x=$ elongation in the spring,

$x=\frac{\mu m_{2}g}{k}....\left(1\right)$

Let the minimum force will be $F$ such that $m_1$ has no change in kinetic energy.

Applying work energy theorem for $m_1$,

${\int }_0^x(F-\mu m_{1}g-kx)dx=0$

$\Rightarrow F[x{]}_0^x-\mu m_{1}g[x{]}_0^x-\frac{k}{2}[x^2{]}_0^x=0$

On solving, we get,

$\Rightarrow F=(\mu m_{1}g+\frac{1}{2}kx)$

$\Rightarrow F=(\mu m_{1}g+\frac{\mu m_{2}g}{2})$

$\Rightarrow F=\mu g(m_1+\frac{m_2}{2})$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (a) is the correct answer.

Key concept Work energy theorem : The theorem states that the total work done on a system equals the change in its kinetic energy.