Question

A student places her $500g$physics book on a frictionless table. She pushes the book against a spring, compressing the spring by $4.0cm$, then releases the book. What is the book’s speed as it slides away? The spring constant is $1250N/m.$

Short answer

book's speed as it slides away on a table$v=2m/s$

Step-by-step solution

Given information

mass of the book $m=500g=0.5Kg$

distance moved by the spring $x=4.0cm=0.04m$

spring constant$K=1250N/m$

Explanation

To estimate the book's speed, the book's kinetic energy and maximum potential energy of spring will be considered.

Book's kinetic energy is given by

$k_e=\frac{1}{2}m{v}^2$

The maximum potential energy of spring is given by

$p_e=\frac{1}{2}K{x}^2$

On equilibrium, we have

role="math" localid="1649914030092" $$\begin{gathered} k_e=p_e \\ \Rightarrow \frac{1}{2}m{v}^2=\frac{1}{2}K{x}^2 \\ \Rightarrow v^2=\frac{K{x}^2}{m} \\ \Rightarrow v=\sqrt{\frac{K{x}^2}{m}} \end{gathered}$$

Plugging the values, we get

$$\begin{gathered} v=\sqrt{\frac{K{x}^2}{m}} \\ =\sqrt{\frac{1250\times {(0.04)}^2}{0.5}} \\ v=2m/s \end{gathered}$$

The book's speed slides on the table is$v=2m/s$