Question

A spring is compressed by 1 cm by a force of 3.92. Then find the potential energy of the spring when it is compressed by 10N ?

Answer 1

The soln is given in the above pic

Answer 2

Answer:

A spring is compressed by 1 cm by a force of 3.92. Then the potential energy of the spring when it is compressed by 10N is $0.12J$

Explanation:

• The elastic potential energy is the energy stored when a force applied to deform an elastic object
• The applied force is proportional to the distance by which the spring is compressed or expanded

        $F=-Kx$ (negative sign means opposite direction)

• 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

compressed distance $x=1cm=0.01m$

the force applied $F=3.92N$

⇒spring constant  $K=F/x$

$K=3.92/0.01=392N/m$

new force applied $F=10N$

the distance compressed due to new force $x=F/K=10/392=0.02m$

substitute these values to get potential energy stored

⇒

$PE=\frac{1}{2} (392*(10/392)^{2} )=0.12J$