Question

A compression spring with a spring constant of 200 N/m catches a ball that's falling with

Answer 1

Answer:

distance of compression: 0.07071 m

Explanation:

$\sf energy = \dfrac{1}{2} kx^2$

$\rightarrow \sf 0.5= \dfrac{1}{2} (200)x^2$

$\rightarrow \sf 0.5= (100)x^2$

$\rightarrow \sf 5\ *\ 10^{-3}= x^2$

$\rightarrow \sf x = \sqrt{5\ *\ 10^{-3}}$

$\rightarrow \sf x =0.07071 \ m$