Physics - W.E.P. Question with Solution | TestHub

Let us take wedge + spring + block as a system. The forces responsible for performing work are spring force kx (←) and the external force F(→).



Work Energy theorem for block + spring + plank relative to ground :

Applying work-energy theorem, we have W_ext+ W_sp= ΔK

where W_spthe total work done by the spring on wedge and block$-\frac{1}{2}\text{k}{x}^2$and ΔK = change in KE of the block (because the plank does not change its kinetic energy)

$\text{Then,}{\text{W}}_{\text{ext}}=\frac{1}{2}\text{k}{x}^2+\text{Δ}\text{K}$

As the block was initially stationary and it will acquire a velocity v₀equal to that of the plank at the time of maximum compression of the spring, the change in kinetic energy of the block relative to ground is

$\text{ Δ}\text{K}=\frac{1}{2}m{v}_0^2$

Substituting ΔK in the above equation, we have

${\text{W}}_{\text{ext}}=\frac{1}{2}\text{K}{x}^2+\frac{1}{2}m{v}_0^2\text{...(i)}$

Work Energy theorem for block + spring + plank  relative to the plank W_ext+ W_sp= ΔK The plank moves with constant velocity, there is no pseudo-force acting on the block. W_ext= 0 Then the net work done on the system (block + plank), due to the spring, can be given as

${\text{W}}_{\text{SP}}=-\frac{1}{2}\text{k}{x}^2$

As the relative velocity between the observer (plank) and block decreases from v₀to zero at the time of maximum compression of the spring, the change in kinetic energy of the block is$\text{Δ}\text{K}=-\frac{1}{2}m{v}_0^2\text{.}$

Substituting W_sp   and  $\stackrel{}{△K}$in above equation
-$\frac{1}{2}k{x}^2= \text{-}\frac{1}{2}m{v}_0^2$(ii)

From (i) & (ii)
${\text{W}}_{\text{ext}}$$=\frac{1}{2}{mv}_0^2+\frac{1}{2}{mv}_0^2={mv}_0^2$