Question

The work done in stretching a spring of constant K by amount by delta​

Answer 1

Answer:

Potential energy in a stretched spring  U=21kx2

where,  x is the extension in the spring.

So, U2=21kl22 and  U1=21kl12

Work done  W=U2−U1

W=21kl22−21kl12=21k(l22−l12).

Answer 2

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

).