Chapter 14.1, Problem 9SSC

To determine

To Compare:The potential energies of two springs.

Answer to Problem 9SSC

The energy of the second spring is fourtimes greater than the energy of first spring.

Explanation of Solution

Given:

Spring 1 is extended by 2 cm and spring 2 is extended by 4 cm.

Formula used:

  $P{E}_{sp}=\frac{1}{2}k{x}^2$

Where,

k: spring constant

x: spring displacement

Calculation:

The potential energy of a spring is given by formula,

  $P{E}_{sp}=\frac{1}{2}k{x}^2$

Consider,

Potential energy of spring

Spring displacement of spring

Potential energy of spring

Spring displacement of spring

To contrast the potential energies of both the springs,

  

  $\frac{P{E}_2}{P{E}_1}=\frac{{0.04}^2}{{0.02}^2}=4$

Conclusion:

Hence, the energy of the second spring is 4 times greater than the energy of first spring.