Fill in the blank: A spring is stretched by 3 | vedclass

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(B) The potential energy $U$ stored in a spring stretched by a distance $y$ is given by $U = \frac{1}{2} k y^2$,where $k$ is the spring constant.For t

Vedclass · Accepted Answer

$4$

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(B) The potential energy $U$ stored in a spring stretched by a distance $y$ is given by $U = \frac{1}{2} k y^2$,where $k$ is the spring constant.For the first case,$y_1 = 3 \, cm$,so $U_1 = \frac{1}{2} k (3)^2 = \frac{9}{2} k$.For the second case,$y_2 = 6 \, cm$,so $U_2 = \frac{1}{2} k (6)^2 = \frac{36}{2} k$.Taking the ratio of the two energies: $\frac{U_2}{U_1} = \frac{\frac{1}{2} k (6)^2}{\frac{1}{2} k (3)^2} = \frac{36}{9} = 4$.Therefore,$U_2 = 4 U_1$.

Fill in the blank: $A$ spring is stretched by $3 \, cm$,its potential energy is $U_1$. If the spring is stretched by $6 \, cm$,its potential energy stored in it is $U_2 = ...... U_1$.

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