Question

A mass of 1 kg is dropped from a height of 2 m on a horizontal spring board. The vertical spring supporting the board has a spring constant of 87.5 N/m. The maximum distance by which the mass compresses the spring is close to

• A. 0.8 m
• B. 0.7 m
• C. 0.6 m
• D. 0.4 m

Answer 1

The correct option is A 0.8 m

We will apply the Principle of Conservation of Mechanical Energy.

We will take the final position of mass $m$ as the reference configuration for gravitational potential energy.

Hence:-

$U_i+K_i=U_f+K_f$

Since, $K_i=K_f=0$

Hence,

$U_i=U_f$

$⟹mg(h+x)=\frac{1}{2}k{x}^2$

$⟹10(2+x)=\frac{87.5}{2}{x}^2$

$⟹87.5x^2-20x-40=0$

$x=\frac{20+\sqrt{{20}^2+4\times 87.5\times 40}}{2\times 87.5}$ neglecting $-ve$ value of x.

$⟹x=0.8m$

Hence, answer is option-(A).