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

Answer:

$X_2 = 0.122m$

Explanation:

Using the law of the convervation of energy E:

$E_i = E_f$

so:

$\frac{1}{2}K_1X_1^2 = \frac{1}{2}K_2X_2^2$

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:

$\frac{1}{2}(553N/m)(0.1m)^2 = \frac{1}{2}(370N/m)X_2^2$

Solving for $X_2$:

$X_2 = 0.122m$