Q143 / 170 · Manipal (MET) 2019 · Physics
When a spring is stretched by 2 cm , it stores 100 J of energy. If it is stretched further by 2 cm , the stored energy will be increased by
- A100 J
- B200 J
- Correct answerC300 J
- D400 J
Solution
Ist case,
$\begin{aligned}
U & =\frac{1}{2} k x_1^2 \\
100 & =\frac{1}{2} k\left(2 \times 10^{-2}\right)^2 \\
k & =\frac{200}{4 \times 10^{-4}} \\
& =50 \times 10^4
\end{aligned}$
IInd case $\begin{aligned} U & =\frac{1}{2} k x_2^2 \\ & =\frac{1}{2} \times 50 \times 10^4\left(4 \times 10^{-2}\right)^2 \\ & =25 \times 10^4 \times 16 \times 10^{-4} \\ & =400 \mathrm{~J} \end{aligned}$
So, stored energy increases by $400-100=300 \mathrm{~J}$
IInd case $\begin{aligned} U & =\frac{1}{2} k x_2^2 \\ & =\frac{1}{2} \times 50 \times 10^4\left(4 \times 10^{-2}\right)^2 \\ & =25 \times 10^4 \times 16 \times 10^{-4} \\ & =400 \mathrm{~J} \end{aligned}$
So, stored energy increases by $400-100=300 \mathrm{~J}$
Hence, the correct answer is option: 'C'.