Question

One end of an elastic spring of natural length $L$ and spring constant $k$ is fixed to a wall and the other end is attached to a block of mass $m$ lying on a horizontal frictionless table. The block is moved to a position A, such that the spring is compressed to half its natural length and then released. What is the velocity of the block when it reaches position B, which is at a distance $\frac{3L}{4}$ from the wall.

• A. $\frac{L}{4}\sqrt{\frac{k}{m}}$
• B. $\frac{L}{2}\sqrt{\frac{2k}{m}}$
• C. $\frac{L}{4}\sqrt{\frac{3k}{m}}$
• D. $\frac{L}{2}\sqrt{\frac{k}{m}}$

Answer 1

The correct option is C $\frac{L}{4}\sqrt{\frac{3k}{m}}$

At position A, the compression of the spring is $\frac{L}{2}$.
At position B, the compression is $L-\frac{3L}{4}=\frac{L}{4}$.
Therefore, loss of potential energy as the block moves from A to B is given by
$\frac{1}{2}k{\left(\frac{L}{2}\right)}^2-\frac{1}{2}k{\left(\frac{L}{4}\right)}^2=\frac{3k{L}^2}{32}$
From conservation of energy, loss in P.E. = gain in K.E.,
i.e.
$\frac{3k{L}^2}{32}=\frac{1}{2}m{v}^2$
$\Rightarrow v=\frac{L}{4}\sqrt{\frac{3k}{m}}$