Question

A horizontal spring with spring constant $250N/m$ is compressed by $12cm$ and then used to launch a $250g$ box across the floor. The coefficient of kinetic friction between the box and the floor is $0.23$. What is the box’s launch speed?

Short answer

The launch speed of the box is$3.7m/s$.

Step-by-step solution

Content Introduction

The work energy theorem is

$$\begin{gathered} W=∆K \\ {F}_{net}.d=∆K \end{gathered}$$

Considering the following diagram,

The force of kinetic friction is

$$\begin{gathered} f_K={\mu }_{K}n \\ {f}_K={\mu }_{K}mg \end{gathered}$$

The net force on the mass is calculated as:

$F_{net}=f_{sp}-f_k$

Therefore, the total work done on the box during the launch is :

$$\begin{gathered} W={\int }_0^{0.12}{F}_{net}dx \\ W={\int }_0^{0.12}(F_{sp}-f_K)dx \\ W=\frac{1}{2}k{x}^2-{\mu }_{K}mgx \\ W=\frac{1}{2}(250N/m){(0.12m)}^2-(0.23)(250g\times \frac{1kg}{1000g})(9.8m/s^2)(0.12m) \\ W=1.73J \end{gathered}$$

Content Explanation

Now apply the work energy theorem,

$$\begin{gathered} W=∆K \\ W=\frac{1}{2}m({v_f}^2-{v_i}^2) \\ 1.73J=\frac{1}{2}m{v_f}^2 \\ {v}_f=\sqrt{\frac{2\times 1.73J}{250g\times {\displaystyle \frac{1kg}{1000g}}}} \\ {v}_f=3.7m/s \end{gathered}$$