BSc Nursing GNM 2019 Solved Paper

Let extension produced in a spring be x initially. In stretched condition, spring will have potential energy,
U=‌

1

2

kx2
where, k is spring constant or force constant.
∴‌‌‌

U1

U2

=‌

x12

x22

. . . (i)
Given,U1=U,x1=2cm,x2=8cm
Putting these values in Eq. (i), we have
‌‌‌

U

U2

=‌

(2)2

(8)2

=‌

4

64

=‌

1

16

∴‌‌U2=16U