Chapter 18, Problem 47A

A certain spring stretches 3 cm when a load of 15 N is suspended from it. How much will the spring stretch if 45 N is suspended from it (and the spring doesn’t reach its elastic limit)?

To determine

The amount of stretch in spring.

Answer to Problem 47A

Spring is stretched by 9 cm.

Explanation of Solution

Given :

When a load of 15 N is suspended spring is stretched by 3 cm.

  $F_1=15\text{ N, }{x}_1=3\text{ cm}$

Second force applied is, $F_2=45\text{ N}$

Formula used :

  $F=kx$

Where, F is force, k is spring constant and x is length of spring stretched.

Calculation :

Now, substituting $F=F_1$ and $x=x_1$ and solve.

  $\begin{array}{l}15=k\times 3\\ k=5\text{ N/cm}\end{array}$

Now, substituting the value of $F=F_2$ and k and solve.

  $\begin{array}{l}45=5\times x_2\\ x_2=9\text{ cm}\end{array}$

Conclusion :

Therefore, spring is stretched by 9 cm.