Question

The natural length of spring is $0.3\text{m}$ and its spring constant is $30\text{N/m}$. Suppose we apply an external force $F$ to stretch the spring. How much work is done by the external force as the spring is stretched from $0.1\text{m}$ to $0.2\text{m}$?

• A. $0.15\text{J}$
• B. $0.3\text{J}$
• C. $0.45\text{J}$
• D. None

Answer 1

The correct option is C $0.45\text{J}$

Given spring constant $\left(k\right)$ = $30\text{N/m}$
Workdone by the deforming force from $x_1=0.1\text{m}$ to $x_2=0.2\text{m}$,
$W=\frac{1}{2}k(x_2^2-x_1^2)=\frac{1}{2}\times 30\times \left(\right(0.2{)}^2-\left(0.1{)}^2\right)=15\times \left(\frac{3}{100}\right)=0.45\text{J}$