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

Step-by-step 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

`(1)/(2)k_1x^2=(1)/(2)mv^2=(1)/(2)k_2x'^2`

`(x')/(x)=\sqrt{(k_1)/(k_2)}`

`x'=0.1* \sqrt{(606)/(233)}`

`x'=0.1* 1.61`

`x'=0.161\ m`