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5.Work, Energy, Power and Collision
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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 ............. $\mathrm{J}$
A$100$
B$200$
C$300$
D$400$
Solution
$100=\frac{1}{2} \mathrm{Kx}^{2}(\text { given })$
$\mathrm{W}=\frac{1}{2} \mathrm{K}\left(\mathrm{x}_{2}^{2}-\mathrm{x}_{1}^{2}\right)=\frac{1}{2} \mathrm{K}\left[(2 \mathrm{x})^{2}-\mathrm{x}^{2}\right]$
$W=3 \times\left(\frac{1}{2} k x^{2}\right)=3 \times 100=300 \mathrm{J}$
$\mathrm{W}=\frac{1}{2} \mathrm{K}\left(\mathrm{x}_{2}^{2}-\mathrm{x}_{1}^{2}\right)=\frac{1}{2} \mathrm{K}\left[(2 \mathrm{x})^{2}-\mathrm{x}^{2}\right]$
$W=3 \times\left(\frac{1}{2} k x^{2}\right)=3 \times 100=300 \mathrm{J}$
Standard 11
Physics
