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 $U = \frac{1}{2} k x^2$,where $k$ is the spring constant and $x$ is the displacement.Initially

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(C) The potential energy stored in a spring is given by $U = \frac{1}{2} k x^2$,where $k$ is the spring constant and $x$ is the displacement.Initially,for $x_1 = 2 \ cm = 0.02 \ m$,the energy $U_1 = 100 \ J$.$100 = \frac{1}{2} k (0.02)^2$$100 = \frac{1}{2} k (4 	imes 10^{-4})$$100 = k (2 	imes 10^{-4})$$k = \frac{100}{2 	imes 10^{-4}} = 5 	imes 10^5 \ N/m$.Now,the spring is further stretched by $2 \ cm$,so the new total displacement is $x_2 = 2 \ cm + 2 \ cm = 4 \ cm = 0.04 \ m$.The new potential energy is $U_2 = \frac{1}{2} k x_2^2$.$U_2 = \frac{1}{2} (5 	imes 10^5) (0.04)^2$$U_2 = \frac{1}{2} (5 	imes 10^5) (16 	imes 10^{-4})$$U_2 = 5 	imes 80 = 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 further stretched by $2 \ cm$,the stored energy will be increased by $.... \ J$.

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