Question

A spring of spring constant $5\times {10}^3\text{N/m}$ is stretched initially by $5cm$ from the unstretched position. The work required to further stretch the spring by another $5cm$ is

• A. $6.25\text{J}$
• B. $12.50\text{J}$
• C. $18.75\text{J}$
• D. $25.00\text{J}$

Answer 1

The correct option is C $18.75\text{J}$

Given spring constant $\left(k\right)=5\times {10}^3\text{N/m}$
Spring force $\left(F\right)=kx$ (where $x$ is deformation in spring)
Work done by deforming force to stretch the spring from $x_1=5cm$ to $x_2=10cm$
$w=\int \overrightarrow{F.}\overrightarrow{ds}={\int }_{x_1}^{x_2}{F}_{x}dx={\int }_{x_1}^{x_2}kxdx=k{\left(\frac{x}{2}\right)}^2{|}_{x_1}^{x_2}=k(\frac{x_2^2}{2}-\frac{x_1^2}{2})=\frac{1}{2}k(x_2^2-x_1^2)=\frac{1}{2}\times 5\times {10}^3\left(\right(10\times {10}^{-2}{)}^2-(5\times {10}^{-2}{)}^2)=\frac{1}{2}\times 5\times {10}^3\times {10}^{-4}(100-25)=\frac{75}{4}=18.75\text{J}$