Question

In the figure shown below, coefficient of friction between the $3\text{kg}$ block and ground varies with time and force acting on the block is given by $F=5t$. Then, work done by the friction force on the $2\text{kg}$ block upto $5\text{sec}$, from the frame of reference of the $3\text{kg}$ block is:
[Take $g=10{\text{m/s}}^2$]

Answer 1

From the figure, we can conclude that at $t=5\text{s}$, external force will be $F=25\text{N}$.
By assuming that both blocks are moving together, let us calculate the friction required between the two blocks.

Friction between the blocks and the ground at $t=5\text{s}$ will be $f_1={\mu }_1\times N=0.02\times 5\times 50=5\text{N}$
and force $F=5t=5\times 5=25\text{N}$
Hence, acceleration of the system will be $a=\frac{F-f_1}{M+m}=\frac{25-5}{5}=4{\text{m/s}}^2.$

For this acceleration, frictional force required on $2\text{kg}$ block is $f_2=2\times 4=8\text{N}$ and maximum possible friction on this block is $f_{2max}={\mu }_2\times mg=0.4\times 20=8\text{N}$.
Thus, both blocks will move together till $t=5\text{s}$ and hence work done by friction between the two blocks will be zero in the frame of reference of $3\text{kg}$ block.
[because there is no relative motion between the blocks]