Question

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

Answer 1

Answer:

The book's speed is $v=3.16 ms^{-1}$.

Explanation:

The expression of the elastic potential energy is as follows;

$PE=\frac{1}{2}kx^{2}$                                              .......... (1)

Here, PE is the elastic potential energy, k is the spring constant and x is the distance.

The expression for kinetic energy is as follows;

$KE=\frac{1}{2}mv^{2}$                                           ............ (2)

Here, m is the mass of the object and v is the speed.

According to the given problem, a student places her 500 g physics book on a frictionless table. She pushes the book against a spring, compressing the spring by 4.0 cm, then releases the book.  The elastic potential energy is converted into kinetic energy.

From (1) and (2),

PE=KE

$\frac{1}{2}kx^{2}=\frac{1}{2}mv^{2}$  

$kx^{2}=mv^{2}$

Rearrange the above expression to calculate the book's speed.

$v=x\sqrt{\frac{k}{m}}$

Convert the distance x from cm to m.

x= 4 cm

x= 0.04 m

Convert the mass m from g to kg.

m= 500 g

m= 0.2 kg

Put x= 0.04 m, m= 0.2 kg and $k=1250 Nm^{-1}$ in the expression for book's speed.

$v=(0.04)\sqrt{\frac{1250}{0.2}}$

$v=3.16 ms^{-1}$

Therefore, the book's speed is $v=3.16 ms^{-1}$.