Question

A 2 kg block is dropped from a height of 0.4m on a spring of force constant $k=1960N{m}^{-1}$ . The maximum compression of the spring is

• A. 0.1 m
• B. 0.2 M
• C. 0.3 M
• D. 0.4 M

Answer 1

Answer: (A)

By law of conservation of energy, we have loss in gravitational potential energy = gain in elastic potential energy $\Rightarrow$ $mg(h+x)=\frac{1}{2}k{x}^2$ $2\times 10(0.4+x)=\frac{1}{2}1960\times x^2$ $\Rightarrow$ $8+20x=980x^2$ $\Rightarrow$ $980x^2-20x-8=0$ Solving, we get $x=0.1m$