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Question

The P.E. of a certain spring certain when stretched from natural length through a distance $$0.3 m $$ is $$10 J$$ . The amount of work in joule that must be done on this spring to stretch it through an additional distance $$0.15 m $$ will be :


A
10J
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B
20J
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C
7.5J
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D
12.5J
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Solution

The correct option is D $$12.5 J$$

The initial potential energy of the spring is given as,

$${U_i} = \dfrac{1}{2}k{x^2}$$

$$10 = \dfrac{1}{2}k{\left( {0.3} \right)^2}$$

$$k = \dfrac{{20}}{{0.09}}$$

The final potential energy of the spring is given as,

$${U_f} = \dfrac{1}{2}k{x_1}^2$$

$$ = \dfrac{1}{2} \times \dfrac{{20}}{{0.09}}{\left( {0.45} \right)^2}$$

$$ = 22.5\;{\rm{J}}$$

The amount of work done is given as,

$$W = {U_f} - {U_i}$$

$$ = 22.5 - 10$$

$$ = 12.5\;{\rm{J}}$$

Thus, the amount of work done is $$12.5\;{\rm{J}}$$.


Physics

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