Question

the elongation of spring is 1 cm and its potential energy is U if the spring is elongated by 3 cm then potential energy will be

Answer 1

potential energy=1/2kx^2

Answer 2

If the spring is elongated by 3 cm then potential energy will be 9U.

Explanation:

Given that,

Elongation in the spring, $x_1=1\ cm=0.01\ m$

Potential energy, $P_1=U$

When the spring is elongated by 3 cm, $x_2=3\ cm=0.03\ m$

We need to find the potential energy when the spring is elongated by 3 cm. The spring potential energy is given by :

$U=\dfrac{1}{2}kx^2$

Initially,

$U=\dfrac{1}{2}kx_1^2$

Let U' is the final potential energy of the spring such that,

$\dfrac{U}{U'}=\dfrac{1/2kx_1^2}{1/2kx_2^2}$

$\dfrac{U}{U'}=\dfrac{x_1^2}{x_2^2}$

$\dfrac{U}{U'}=\dfrac{(0.01)^2}{(0.03)^2}$

$\dfrac{U}{U'}=0.112$

U' = 9U

So, if the spring is elongated by 3 cm then potential energy will be 9U. Hence, this is the required solution.

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