Work Energy Power

JEE Main level, single correct, medium

139 of 183

Question 139 of 183

A spring of spring constant 5×103 N/m5 \times 10^3 \mathrm{~N} / \mathrm{m} is stretched initially by 5 cm5 \mathrm{~cm} from the unstretched position. Then the work required to stretch it further by another 5 cm5 \mathrm{~cm} is

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Answer: (B) 18.75 N−m18.75 \mathrm{~N}-\mathrm{m}

Required work done
=12 K(x22−x12)=12×5×103[102−52]×10−4=\frac{1}{2} \mathrm{~K}\left(\mathrm{x}_2^2-\mathrm{x}_1^2\right)=\frac{1}{2} \times 5 \times 10^3\left[10^2-5^2\right] \times 10^{-4}
=12×5×75×103×10−4=18.75=\frac{1}{2} \times 5 \times 75 \times 10^3 \times 10^{-4}=18.75

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