Question

A spring of spring constant 5 x ${10}^3$ N/m is etched initially by 5 cm from the unstretched position the work required to stretch it further by another 5 cm is

• A. $18.75N-m$
• B. $25.00N-m$
• C. $6.25N^{-}m$
• D. $12.50N^{}-m$

Answer 1

The correct option is A $18.75N-m$

Work done is change in potential energy

The spring is initially stretch to 5 cm and again further stretched by 5 cm and

further stretched by 5 cm

$\begin{array}{c}w=p_{10}-p_5\\ =\frac{1}{2}\times 5\times {10}^3\left[I{\left(\frac{10}{100}\right)}^2-{\left(\frac{5}{100}\right)}^2\right]\\ =\frac{1}{2}\times 5\times {10}^3\left(\frac{10}{100}+\frac{5}{100}\right)\left(\frac{10}{100}-\frac{5}{100}\right)N-m\\ =\frac{1}{2}\times 5\times {10}^3\times \frac{15}{100}\times \frac{5}{100}=2.5\times 15\times 5\times {10}^3\times {10}^{-4}\\ =75\times 2.5\times {10}^{-1}=7.5\times 2.5N-m\\ =18.75N-m\end{array}$

Option A.