Question

A spring of negligible mass has force constantk = 800 N / m . (a) How far must the spring be compressed for 1.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then lay a 1.60 kg book on top of the spring and release the book from rest. Find the maximum distance the spring will be compressed.

Short answer

1. The compression length that must the spring be compressed for 1.20 J of potential energy to be stored in it is 5.48 cm.
2. The maximum distance the spring will be compressed is 1.96 cm.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The force constant of a spring is, k = 800 N/m .
• The amount of potential energy stored in the spring is, PE = 1.20 J .
• The mass of a book is, m = 1.60 kg .

Significance of gravitational potential energy

When a mechanical spring is subjected to an external force, there would be a force generated by the spring that is referred to as spring force. The value of the spring force helps to obtain the value of stored energy in the spring.

(a) Determination of the compression length that must the spring be compressed for of potential energy to be stored in it

The relation to calculate the compression length that must the spring be compressed for 1.20J of potential energy to be stored in it is expressed as,

$$\begin{gathered} \left(\mathrm{PE}\right)=\frac{1}{2}\mathrm{k}{\left(∆\mathrm{x}\right)}^2 \\ ∆\mathrm{x}=\sqrt{\frac{2\left(\mathrm{PE}\right)}{\mathrm{k}}} \end{gathered}$$

Here, $∆\mathrm{x}$is the compression length that must the spring be compressed for 1.20 J of potential energy to be stored in it.

Substitute all the known values in the above equation.

$$\begin{gathered} ∆\mathrm{x}=\sqrt{\left\{\frac{2\left(1.20\mathrm{J}\right)}{800\mathrm{N}/\mathrm{m}}\right\}\left\{\frac{1\mathrm{m}}{1\mathrm{J}.\mathrm{m}/\mathrm{N}}\right\}} \\ \approx 0.0548\mathrm{m} \\ \approx \left(0.0548\mathrm{m}\right)\times \left(\frac{{10}^2\mathrm{cm}}{1\mathrm{m}}\right) \\ \approx 5.48\mathrm{cm} \end{gathered}$$

Thus, the compression length that must the spring be compressed for 1.20 J of potential energy to be stored in it is 5.48 cm.

(b) Determination of the maximum distance the spring will be compressed

The relation ofto calculate themaximum distance the spring will be compressedis expressed as,

$$\begin{gathered} \mathrm{W}={\mathrm{F}}_{\mathrm{s}} \\ \mathrm{mg}=\mathrm{k}∆\mathrm{s}\text{'} \\ ∆\mathrm{x}\text{'}=\left(\frac{\mathrm{mg}}{\mathrm{k}}\right) \end{gathered}$$

Here, $∆\mathrm{x}\text{'}$is the maximum distance the spring will be compressed, W is the weight of the book, ${\mathrm{F}}_{\mathrm{s}}$is the spring force and is the gravitational acceleration whose value is 9.81$\mathrm{m}/{\mathrm{s}}^2$.

Substitute all the known values in the above equation.

$$\begin{gathered} ∆\mathrm{x}\text{'}=\left[\frac{\left(1.60\mathrm{kg}\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}{800\mathrm{N}/\mathrm{m}}\left(\frac{1\mathrm{m}}{1\mathrm{kg}.{\mathrm{m}}^2/\mathrm{N}.{\mathrm{s}}^2}\right)\right] \\ =0.01962\mathrm{m} \\ =\left(0.01962\mathrm{m}\right)\times \left(\frac{{10}^2\mathrm{cm}}{1\mathrm{m}}\right) \\ \approx 1.96\mathrm{cm} \end{gathered}$$

Thus, the maximum distance the spring will be compressed is 1.96 cm.