Question

When 1 N force is applied increase in length of the spring is 1 cm. Find elastic potential energy stored during

this in it.

A) 10 × 10-3

J B) 10-3

J C) 5× 10-3

J D) 20 × 10-3

J​

Answer 1

Explanation:

Answer is (C) 5x10-3

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Answer 2

Answer:

When 1 N force is applied increase in length of the spring is 1 cm, the elastic potential energy stored is $5*10^{-3} J$

Explanation:

• The elastic potential energy is the energy stored when a force applied to deform an elastic object
• From Hook's law the restoring force in spring with spring constant K when its natural length changed to x  is $F=-Kx$

• A spring with elastic constant K compressed or expanded at a distance x from its mean position, then the potential energy stored is

        $PE=\frac{1}{2}Kx^{2}$

From the question, we have

the applied force is $F=1N$

increase in length of the spring $x=1cm=0.01m$

the spring constant of the spring $K=1/0.01=100N/m$

⇒

potential energy stored $PE=\frac{1}{2} (100*0.01^{2} )=5*10^{-3}J$