Question

A small ball of mass M is held in equilibrium with two identical springs as shown in the figure. Force constant of each spring is k and relaxed length of each spring is $l/2$. What is distance between the ball and roof?

• A. $\frac{l}{2}+\frac{Mg}{k}$
• B. $\frac{l}{2}-\frac{Mg}{k}$
• C. $\frac{l}{2}+\frac{Mg}{2k}$
• D. $\frac{l}{2}-\frac{Mg}{2k}$

Answer 1

The correct option is C $\frac{l}{2}+\frac{Mg}{2k}$

The first spring is stretched by x and the second spring is compressed be X

Net force acting on the ball is 0

Mg=2kx

$x=\frac{Mg}{2k}$

Distance between ball and roof is $\frac{l}{2}+x=\frac{l}{2}+\frac{Mg}{2k}$