Question

A block of mass m = 0.23 kg is set against a spring with a spring constant of k1 = 553 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 = 370 N/m. How far d2, in meters, will the second spring compress when the block runs into it?

Answer 1

`X_2 = 0.122m`

Step-by-step explanation:

Using the law of the convervation of energy E:

`E_i = E_f`

so:

`(1)/(2)K_1X_1^2 = (1)/(2)K_2X_2^2`

where
`K_1`is the constant of the spring 1,
`X_1` the compressed of the first spring,
`K_2` is the constant of the second spring and
`X_2` is the compressed of the second spring.

Replacing values, we get:

`(1)/(2)(553N/m)(0.1m)^2 = (1)/(2)(370N/m)X_2^2`

Solving for
`X_2`:

`X_2 = 0.122m`