Question

A horizontal spring with spring constant$85$ N/m extends outward from a wall just above floor level. A $1.5$kg box sliding across a frictionless floor hits the end of the spring and compresses it $6.5$cm before the spring expands and shoots the box back out. How fast was the box going when it hit the spring?

Short answer

The initial velocity of the box with it hit the spring is $0.49$m/s

Step-by-step solution

Step 1. Given information

Spring constant$k=84N/m$

Mass of box$m=1.5kg$

Length by which spring compressed$∆s_f=6.5cm=0.065m$

Step 2. Find the speed

The speed with which the box hit the spring

$$\begin{gathered} W=\left(\frac{1}{2}k∆{s_f}^2\right)+\left(\frac{1}{2}k∆{s_i}^2\right) \\ =-\frac{1}{2}\times 85\times 0.{065}^2+0 \\ =-0.1795J \\ ∆K=\frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2 \\ ∆K=0-\frac{1}{2}m{v_i}^2 \\ W=∆K \\ \Rightarrow -\frac{1}{2}m{v_i}^2=-0.1795 \\ {v}_i=\sqrt{\frac{0.1795\times 2}{1.5}}=0.49m/s \end{gathered}$$