Question

If a spring of spring constant 10N/cm is compressed by 2cm.Then it's potential energy is​

Answer 1

Answer:

EPE = 1/2 K E

$\frac{1}{2} \times 0.1 \times 0.02$

0.004J

10N/cm = 0.1N/m

Answer 2

Answer:

If a spring of spring constant 10N/cm is compressed by 2cm. Then it's potential energy is​ $0.2J$

Explanation:

• The elastic potential energy is the energy stored when a force applied to deform an elastic object

• 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

spring constant $K=10N/cm=1000N/m$

compressed distance $x=2cm=0.02m$

substitute these values to get potential energy stored

⇒

$PE=\frac{1}{2} (1000*(0.02)^{2} )=0.2J$