Chapter 11, Problem 2P

To determine

To Find:The value of spring constant.

Answer to Problem 2P

The value of spring constant is $k=\text{524}\text{.4 N/m}$ .

Explanation of Solution

Given:

Forces on elastic cord, $F_1=75$ N, $F_2=180$ N.

Extension in cord ${\text{L+ x}}_1=0.65\text{m}$ , ${\text{L+ x}}_2=0.85\text{m}$ .

Where L is length of upstretched cord .

Formula used:

Spring force

  $F=-kx$ .

Here, k is the spring constant and x is the extension in cord.

Calculation:

Using above equations

  $F_1=-k{x}_1,F_2=-k{x}_2$-----(1)

  $F_1=-k(0.65-L),F_2=-k(0.85-L)$

Dividing these two

  $\frac{F_1}{F_2}=\frac{-k(0.65-L)}{-k(0.85-L)}$

  $\frac{75}{180}=\frac{-k(0.65-L)}{-k(0.85-L)}$

  $\text{L = }\frac{\text{(0}\text{.85}\times \text{75+0}\text{.65}\times \text{180)}}{105}$

  $\text{L = 0}\text{.507 m}$ .

Using value of L in (1)

  $75=-k(0.65-0.507)$

  $k=\text{524}\text{.4 N/m}$ .

Conclusion:

Thus, the value of spring constant is $k=\text{524}\text{.4 N/m}$ .