Chapter 14.1, Problem 4PP

To determine

Displacement covered by the stretching spring to have an elastic potential energy of 48 J.

Answer to Problem 4PP

  $61cm$

Explanation of Solution

Given info:

Potential energy $=48J$

Spring constant $=256N/m$

Formula used:

The potential energy stored in a spring

  $P{E}_{sp}=\frac{1}{2}k{x}^2$

Here, k is the spring constant and x is the stretch in the spring.

Calculation:

As per given information,

Here, $k$

  $=256N/m$

  $P{E}_{sp}$

  $=48J$

Now, substitute the values in

  $P{E}_{sp}=\frac{1}{2}k{x}^2$

  $48=\frac{1}{2}\left(256\right)x²$

  $48=128x²$

Divide both sides by $128$ and take square root

  $\sqrt{\frac{48}{128}}=x$

  $x$

  $=\sqrt{\frac{48}{128}}\approx 0.61m=61cm$

Conclusion:

Thus, spring will be stretched $61cm$ far.