A $4\text{kg}$ block is placed on top of a long $12\text{kg}$ block, which is accelerating along a smooth horizontal table at $a=5.2\text{m}/{\text{s}}^2$ under application of an external constant force. let minimum co-efficient of friction between the two blocks which will prevent the $4\text{kg}$ block from sliding is $\mu$ and co-efficient of friction between the blocks is only half of this minimum value i.e., $\frac{\mu }{2}$. Find the amount of heat generated due to sliding between the two blocks during the time in which $12\text{kg}$ block moves $10\text{m}$ starting from rest.

The correct options are
C $52\text{J}$
G $52\text{J}$
First let us assume that both the blocks are moving with common accelaration i.e., $a=5.2\text{m}/{\text{s}}^2$


For $4\text{k}g$ block:

$f=ma$

${\mu }_{0}mg=m\left(5.2\right)$

$\therefore {\mu }_0=0.52$

Given, $\mu =\frac{{\mu }_0}{2}=\frac{1}{2}\times 0.52=0.26$,

The acceleration of $4\text{k}g$ block is due to friction.

$a_1=\mu g=2.6\text{m}/{\text{s}}^2$

As there is relative motion between the blocks, we apply $S_{rel}$,

$S_{rel}=u_{rel}+\frac{1}{2}{a}_{rel}.t^2$

$a_{rel}=a_1-a=2.6-5.2=-2.6\text{m}/{\text{s}}^2$

$\therefore S_{rel}=\frac{1}{2}\times (-2.6)\times t^2......\left(1\right)$

Time of motion can be determined from motion of lower block.

For $12\text{kg}$ block,

$S=\frac{1}{2}\times \left(5.2\right)\times t^2$

$10=\frac{1}{2}\times \left(5.2\right)\times t^2$

$\therefore t\approx 2\text{s}$

Substituting $t$ in eq(1) we get,

$\therefore S_{rel}\approx -5\text{m}$

Work done by friction is given by,

$W_f=\mu mg{S}_{rel}=0.26\times 4\times 10\times (-5)$

$W_f=-52\text{J}$

$\therefore$ Heat generated = $52\text{J}$

Hence , option (c) is the correct answer.