Question

A cube of mass m = 0.49 kg is set against a spring with a spring constant of k1 = 606 N/m which has been compressed by a distance of 0.1 m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k2 = 233 N/m. How far d2, in meters, will the second spring compress when the block runs into it?

Answer 1

Answer:0.161 m

Explanation:

Given

mass of cube  $m=0.49\ kg$

Spring constant $k_1=606\ N/m$

compression in the spring $x_1=0.1\ m$

When this cube is released then it will compress another spring of spring constant $k_2=233\ N/m$

Conserving energy

$\frac{1}{2}k_1x^2=\frac{1}{2}mv^2=\frac{1}{2}k_2x'^2$

$\frac{x'}{x}=\sqrt{\frac{k_1}{k_2}}$

$x'=0.1\times \sqrt{\frac{606}{233}}$

$x'=0.1\times 1.61$

$x'=0.161\ m$