Question

A spring of force constant 'k' is stretched by a small length '//'. The work done in stretching it further by a small length 'ly' is,
(A) (1) + 1%)
(B) , k(1, +1,)?
(C) , kl» (211 – 12)
(D) ,k(13 - 1)​

Answer 1

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

).

Answer 2

Explanation:

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