Question

The spring shown in figure is unstretched when a man starts pulling the block. The mass of the block is $M$. If the man exerts a constant force $F$.
The energy stored in the spring when the block passes through the equilibrium position is

• A. $\frac{2F^2}{k}$
• B. $\frac{F^2}{k}$
• C. $\frac{F^2}{4k}$
• D. $\frac{F^2}{2k}$

Answer 1

Answer: (D)

$\frac{F^2}{2k}$

$F=kx$

Energy in the spring when $x=\frac{F}{k}$ is $=\frac{1}{2}k(\frac{F}{k}{)}^2$

$=\frac{F^2}{2k}$