Question

spring constant of a spring is 5 dyne cm-1 and it is elongated from 20 cm to 22 cm the energy stored in the spring is​

Answer 1

Energy stored in the spring will be $10^{-6}J$

Explanation:

We have given spring constant of the spring K = 5 dyne/cm

We know that $1dyne=10^{-5}N$

And 1 cm = 0.01 m

So $5dyne/cm=5\times \frac{10^{-5}N}{0.01m}=5\times 10^{-3}N/m$

It is given that spring elongated from 20 cm to 22 cm

So change in length of the spring x = 22 - 20 = 2 cm

As 1 cm = 0.01 m

So 2 cm = 0.02 m

Now energy stored in the spring will be equal to $E=\frac{1}{2}Kx^2$

So $E=\frac{1}{2}\times 5\times 10^{-3}\times 0.02^2=10^{-6}J$

So energy stored in the spring will be $10^{-6}J$

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Answer 2

Answer:

10 erg

Explanation:

Hope it helps you.........