Q 11-05-213JEE MainMedium
A block of mass $1$ kg is dropped from a height of $40$ cm onto the top of a light vertical spring of force constant $100$ N/m fixed on the ground. The maximum compression of the spring is ______ cm. (Take $g = 10\ \text{m/s}^2$)
Numerical value type. Enter your answer.
Answer: 40
Concept. At maximum compression the block is momentarily at rest. All the gravitational energy lost over the fall $h + x$ is stored in the spring.
Step 1. Energy equation.
$$mg(h + x) = \tfrac12kx^2 \;\Rightarrow\; 10(0.4 + x) = 50x^2$$
Step 2. Solve.
$$25x^2 - 5x - 2 = 0 \;\Rightarrow\; x = \frac{5 + \sqrt{25 + 200}}{50} = \frac{20}{50} = 0.4\ \text{m}$$
Answer. $40$.
Solution by Sreeraj P, M.Sc Physics