Chapter 18, Problem 48A

If a certain spring stretches 4 cm when a load of 10 N is suspended from it, how much will the spring stretch if it is cut in half and 10 N is suspended from it?

To determine

The amount of stretch in spring.

Answer to Problem 48A

Spring is stretched by 2 cm.

Explanation of Solution

Given :

When a load of 10 N is suspended spring is stretched by 3 cm.

  $F_1=10\text{ N, }{x}_1=4\text{ cm}$

Second force applied is, $F_2=10\text{ N}$

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.

  $\begin{array}{l}10=k\times 4\\ k=\frac{10}{4}\\ k=2.5\text{ N/cm}\end{array}$

As the spring cut into two parts and spring constant is inversely proportional to length of spring therefore spring constant will be double for each part.

Therefore, new spring constant.

  $\begin{array}{l}{k}^{\text{'}}=2k\\ =2\times 2.5\\ =5\text{ N/cm}\end{array}$

Now, substituting the value of $F=F_2$ and $k^{\text{'}}$ and solve.

  $\begin{array}{l}10=5\times x_2\\ x_2=\frac{10}{5}\\ x_2=2\text{ cm}\end{array}$

Conclusion :

Therefore, spring is stretched by 2 cm.