Question

The work done in streching a spring of force constant from lenght L1 and L2 is
​

Answer 1

Explanation:

ANSWER

Potential energy in a stretched spring U=

2

1

kx

2

where, x is the extension in the spring.

So, U

2

=

2

1

kl

2

2

and U

1

=

2

1

kl

1

2

Work done W=U

2

−U

1

W=

2

1

kl

2

2

−

2

1

kl

1

2

=

2

1

k(l

2

2

−l

1

2

).