Question

the length of a spring is increased by 2.5 cm the elastic potential energy stored in the spring is U if the length is increased by 5cm its elastic potential energy become​

Answer 1

Explanation:

The potential energy of a stretched spring is

U=21kx2

Here, k= spring constant,

          x= elongation in spring.

But given that, the elongation is 2cm.

So,   U=21k(2)2

⇒U=21×4         (i)

If elongation is 10cm then potential energy

U′=21k(10)2

U′=21k×100       (ii)

On dividing equation (ii) by equation (i), we have

UU′=21k×421k×100

or    UU =25⇒U =25U