Question

Find the maximum compression in the spring, if the lower block is shifted rightwards with acceleration 'a'. All the surfaces are smooth

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Method : 1 :- Without Energy

Freebody drawing with respect lower block

Now ma' = ma - kg (a' is the net accelearation of the block with respect to lower block)

$\Rightarrow$ = mv $\frac{dv}{dx}$ = ma - kz

$\Rightarrow$ mvdv = (ma - kx)dx

Integrating ,

$\underset{0}{\overset{0}{\int }}$MUdv = $\underset{0}{\overset{x}{\int }}amax$ (Ma - kx)dx

Intial and final velocity will be zero

$\therefore$ 0 = $ma{x}_{max}$ - $\frac{k{x}_{{max}^{max}}}{r}$

$\Rightarrow$ $x_{max}$ = $\frac{2ma}{k}$