Q. A spring stores 1 J of energy for compression of 1 mm. The additional work to be done to compress it further by 1 mm is
1) 1 J
2) 2 J
3) 3 J
4) 4 J
5) 0.5 J
Answer:
3) 3 J
Solution:
W1 = (1/2)kx2
= (1/2)k(1 x 10-3)2
k = 2 x 106
W2 = (1/2)(2 x 106) (1 x 10-3 + 1 x 10-3)2
= 4 J
The additional work done = W2 – W1
= 4J – 1J = 3J