When a spring is stretched by 2 cm,it stores | vedclass

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(C) The potential energy stored in a spring is given by $U = \frac{1}{2} k x^2$.Initially,for $x_1 = 2 \ cm$,the energy $U_1 = 100 \ J$.So,$100 = \fra

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$300$

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(C) The potential energy stored in a spring is given by $U = \frac{1}{2} k x^2$.Initially,for $x_1 = 2 \ cm$,the energy $U_1 = 100 \ J$.So,$100 = \frac{1}{2} k (2)^2 \implies 100 = 2k \implies k = 50 \ J/cm^2$.When stretched further by $2 \ cm$,the new extension is $x_2 = 2 \ cm + 2 \ cm = 4 \ cm$.The new potential energy is $U_2 = \frac{1}{2} k x_2^2 = \frac{1}{2} (50) (4)^2 = 25 	imes 16 = 400 \ J$.The increase in stored energy is $\Delta U = U_2 - U_1 = 400 \ J - 100 \ J = 300 \ J$.

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 ............. $J$.

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