A force of F = 0.5 N is applied on lower block as shown in figure . The work done by lower block on upper block for a displacement of 3 m of the upper block with respect to ground is (Take, 9 = 10ms ²)

Maximum acceleration of I kg block,   $a_{\max }=\mu g=1{ \mathrm{ms}}^{-2}$

Common acceleration without relative motion between two blocks,, $a=\frac{0.5}{3}{ \mathrm{ms}}^{-2}$

Since,  $a<a_{\max }$

There will be no relative motion and blocks will move with  acceleratlon  $\frac{0.5}{3}{ \mathrm{ms}}^{-2}$

Force of friction by lower block on upper block

$f=ma=\left(1\right)\left(\frac{0.5}{3}\right)=\frac{1}{6} \mathrm{N}\text{ (towards right) }$

Work done,   $H=f\times s=0.5\mathrm{J}$