Question

A spring of spring constant 200 Nm-1 is compressed
by 2 cm. The potential energy stored in spring is
(1) 0.01 J
(2) 0.02 J
(3) 0.03 J
(4) 0.04 J​

Answer 1

K=200N/m

x=2cm = 0.02m

Formula for potential energy stored in a spring :

U=1/2 kx^2

U=1/2 (200)(0.02)^2

U=0.04J

Answer 2

Explanation:

answer for this question