Question

A spring With spring constant k when compressed by 1 cm by a force of 4N. Find the potential energy of the spring when it is compressed by 10 cm

Answer 1

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Answer 2

Answer:

A spring With spring constant k when compressed by 1 cm by a force of 4N, the potential energy of the spring when it is compressed by 10 cm $2J$

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 $F=4N$

⇒spring constant  $K=F/x$

$K=4/0.01=400N/m$

new distance of compression $x=10cm=0.1m$

substitute these values to get potential energy stored

⇒

$PE=\frac{1}{2} (400*0.1^{2} )=2J$