Question

When a spring is stretched through a distance of 2 cm its potential energy is 1.5 the work required to stretch it for the distance of 6 cm will be?

Answer 1

Answer:

The work required to stretch it for the distance of 6 cm will be 4.5 Joule

Explanation:

If the spring constant is $k$ and it is displaced by $x$ then the potential energy in the spring is given by

$PE=\frac{1}{2}kx^2$

Therefore

$1.5=\frac{1}{2}k\times 2^2$

$\implies k=\frac{1.5}{2}$

When it is stretched by 6 cm the Potential Energy

$PE=\frac{1}{2}k\times 6^2$

$\implies PE=\frac{1}{2}\times\frac{1.5}{2} \times 6^2$

$\implies PE=\frac{1}{2}\times 1.5 \times 6$

$\implies PE=4.5$ J

Therefore, the work required to stretch it for the distance of 6 cm will be 4.5 Joule

Hope this helps.