Question

A 2.0 kg box that is sliding on a frictionless surface with a speed of 13 m/s approaches a horizontal spring. (See Fig. 5.18.) The spring has a spring constant of 2000 N/m.
(a) How far will the spring be compressed in stopping the box?
(b)In order to reduce the boxes speed to half of its initial speed, how far must the spring be compressed?

Answer 1

(a)
Let's we analyze the situation, the box is moving, so there must be some kinetic energy, then all the kinetic energy is transformed to spring potential energy.
There is no friction, so everything is conservative.

\[\frac{1}{2}mv^2 = \frac{1}{2}kx^2\]

We want to find x.

\[x = \sqrt{\frac{mv^2}{k}}\]