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
B
20J
C
7.5J
D
12.5J

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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