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 $U = \frac{1}{2}kx^2$.Initially,for $x_1 = 2 \ cm$,the energy $U_1 = 100 \ J = \frac{1}{2}k(2)

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(C) The potential energy stored in a spring is given by $U = \frac{1}{2}kx^2$.Initially,for $x_1 = 2 \ cm$,the energy $U_1 = 100 \ J = \frac{1}{2}k(2)^2$.When the spring is stretched by another $2 \ cm$,the total extension becomes $x_2 = 2 \ cm + 2 \ cm = 4 \ cm$.The total energy stored at $x_2$ is $U_2 = \frac{1}{2}k(4)^2 = \frac{1}{2}k(16)$.Since $U_1 = \frac{1}{2}k(4) = 100 \ J$,we have $\frac{1}{2}k = 25 \ J/cm^2$.Therefore,$U_2 = 25 	imes 16 = 400 \ J$.The additional energy stored 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 by another $2 \ cm$,the additional energy stored is ....... $J$.

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