Question

A plank of mass $m_1=8kg$ with a bar of mass $m_2=2kg$ placed on its rough surface, lie on smooth floor of elevator ascending with an acceleration $\frac{g}{4}$. The coefficient of friction is $\mu =\frac{1}{5}$ between $m_1$ and $m_2$. A horizontal force $F=30N$ is applied to the plank. Then the acceleration of bar and the plank in the reference frame of elevator are: ($g=10{\text{m/s}}^2)$

• A. $3.5m/s^2,0.5m/s^2$
• B. $5m/s^2,\frac{30}{8}m/s^2$
• C. $2.5m/s^2,\frac{25}{8}m/s^2$
• D. $4.5m/s^2,4.5m/s^2$

Answer 1

The correct option is C $2.5m/s^2,\frac{25}{8}m/s^2$

Looking from the frame of reference of the lift, a psuedo force will act on the block and the plank in the downward direction.
Limiting friction at the contact surface of block and plank $=\mu (m_{2}g+m_2\left(\frac{g}{4}\right))=0.2\times 25N=5N$
Now let us assume that friction sufficient to ensure no slipping occurs, be ${}^{\prime }{f}^{\prime }$.
So acceleration for both blocks will be equal, say ${}^{\prime }{a}^{\prime }$.
Now, second law in horizontal direction for both masses gives,
$F-f=8a$
and $f=2a$
Thus,
$a=3\frac{m}{s^2}$
$f=6N$
But we know that the limiting value of friction is $5N$. This means that friction is not large enough to prevent slipping. Hence, the block and the plank will slide on each other. Hence, both will have unequal accelerations. And friction in this case is sliding friction $=5N$.
Thus we get,
$F-5=8a_1$
and $5=2a_2$
Thus, $a_2=2.5\frac{m}{s^2}$ & $a_1=\frac{25}{8}\frac{m}{s^2}$