Wave Mechanics Question 514
Question: When a 4 kg mass is hung vertically on a light spring that obeys Hooke’s law, the spring stretches by 2cms. The work required to be done by an external agent in stretching this spring by 5cms will be $ (g=9.8m/sec^{2}) $
Options:
A) 4.900joule
B) 2.450joule
C) 0.495 joule
D) 0.245 joule
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ K=\frac{F}{x}=\frac{4\times 9.8}{2\times {{10}^{-2}}}=19.6\times 10^{2} $ Work done $ =\frac{1}{2}19.6\times 10^{2}\times {{(0.05)}^{2}}=2.45J $
BETA
