Question

Two discs each having mass m are attached rigidly to ends of spring. One of the discs rests on a horizontal surface and the other produces a compression x on the spring when it is in equilibrium. How much further must the spring be compressed so that when the force causing the compression is removed the extension of the spring will be able to lift the lower disc off the table?

• A. $\sqrt{2}x$
• B. $2$x
• C. $1.5$x
• D. $3$x

Answer 1

The correct option is B $2$x

so both the discs are identical, force require to lift the lower disc equal to the weight of the upper disc. i.e. same extension as compression of the spring.

so we require is minimum of $x$ expansion in the spring to lift the lower disc.

when upper disc is released system contains only potential energy in spring, then potential energy of spring is converted to kinetic energy of upper disc and then finally potential energy of spring and potential energy of upper disc, what we require is at some point there must be some kinetic energy in the upper disc when force acting on the lower disc is equal to its weigh. i.e. expansion is $x$.

let spring is further compressed by $p$.

so conserving energy $\frac{1}{2}k(p+x{)}^2=\frac{1}{2}k{x}^2+mg(p+2x)$

also we know $kx=mg$

so we get $\frac{1}{2}k(p+x{)}^2=\frac{1}{2}k{x}^2+kx(p+2x)$

$p=2x$