Question

30. Derive an expression for the potential energy stored in a
system of a block attached to a massless spring, when the
block is pulled from its mean position.
(H.P.S.S.C.E. 2010, 2007)​

Answer 1

Answer:

U=1/2(K x X^2)

Explanation:

Answer 2

The derivation is given below .

We know , work done is given by :

$W=F.dx$     ......( 1 )

Now , in simple harmonic motion force F is given by :

$F=-kx$

Work done in moving object by small displacement dx is :

$dW=-kx.dx$

Therefore ,

$W=-\int\limits^x_0 {kx} \, dx \\\\W=-k\int\limits^x_0 {x} \, dx \\\\W=-k[\dfrac{x^2}{2}]_0^ x\\\\W=-\dfrac{kx^2}{2}$

Now , by work theorem .

P= - W

Therefore ,

$\text{Potential energy}=\dfrac{kx^2}{2}$

Hence , this is the required solution .

Learn More :

Simple Harmonic Motion

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