Question

(3) 300 J
41.
(4) 400 J
33. The potential energy of a certain spring when stretched
through a distance s is 10 Joule. The amount of work
(in joule) that must be done on this spring to stretch it
through an additional distance s will be
[CPMT 2002]
@ (1) 30
(2) 40
(4) 20
(3) 10​

Answer 1

Answer:

We know that Potential energy is u=

2

1

Ks

2

So, on substituting we have ⇒ 10=

2

1

Ks

2

Workdone, W=

2

1

K(2s)

2

−

2

1

Ks

2

W=2Ks

2

−

2

1

Ks

2

=

2

3

Ks

2

Thus, W=3×10=30 Joules