Question

A $50g$rock is placed in a slingshot and the rubber band is stretched. The magnitude of the force of the rubber band on the rock is shown by the graph in FIGURE $P9.55$. The rubber band is stretched $30cm$and then released. What is the speed of the rock?

Short answer

The speed of the rock is$19m/s$

Step-by-step solution

Content Introduction

The formula to find the force due to rubber band is

$F_{sp}=k∆x$

Here, $F_{sp}$is the force due to rubber, $k$is the force constant of rubber, $∆x$is the change in the length of the rubber. Consider the below graph

For $∆x=10cm$the force is $20N$.

Therefore, the force constant is calculated as,

$$\begin{gathered} F_{sp}=k∆x \\ k=\frac{F_{sp}}{∆x} \\ k=\frac{20N}{10cm{\displaystyle \frac{1m}{100cm}}} \\ k=200N/m \end{gathered}$$

Content Explanation

The conservation of energy for the rock is

$$\begin{gathered} K_f+U_{s,f}=K_i+U_{s,i} \\ \frac{1}{2}m{v_f}^2+\frac{1}{2}k{x_f}^2=\frac{1}{2}m{v_i}^2+\frac{1}{2}k{x_i}^2 \\ {v}_f=x_i\times \sqrt{\frac{k}{m}} \\ {v}_f=(30cm\times \frac{1m}{100cm})\times \sqrt{\frac{200N/m}{50g\times {\displaystyle \frac{1kg}{1000g}}}} \\ {v}_f=19m/s \end{gathered}$$