Physics - NLM Question with Solution | TestHub

For $t=0$ to 1 sec:

$$a_T=5\mathrm{m}/{\mathrm{s}}^2$$

Now, in the frame of the truck:

$$ma-\mu mg=m{a}_{B/T}$$

$$a_{B/T}=a-\mu g=5-0.2\times 10=3\mathrm{m}/{\mathrm{s}}^2$$

Now,

$$a_{B/T}=a_B-a_T$$

$$a_B=a_{B/T}+a_T=3-5=-2\mathrm{m}/{\mathrm{s}}^2$$

Thus, $a_B=-2\mathrm{m}/{\mathrm{s}}^2$.

Now, slipping will stop when both the block and the truck have the same velocity. The velocity of the truck after 1s is:

$$v=u+at$$

$$v=0+(-5)\times 1\Rightarrow v=-5\mathrm{m}/\mathrm{s}$$

Thus, the block should have a velocity of $-5\mathrm{m}/\mathrm{s}$, so:

$$v=u+at$$

$$-5=0+(-2)\times t$$

$$t=2.5\sec$$

Thus, after $t=2.5\sec$, slipping will stop. Considering the direction of the truck's acceleration as positive, the velocity of the block with respect to the ground is plotted as shown in option (C).