Physics Elasticity Question 66
When a 4 kg mass is hung vertically on a light spring that obeys Hooke’s law, the spring stretches by 2 cms. The work required to be done by an external agent in stretching this spring by 5 cms will be $ (g=9.8,m/s^{2}) $
[MP PMT 1995]
Options:
A) 4.900 joule
B) 2.450 joule
C) 0.495 joule
D) 0.245 joule
Show Answer
Answer:
Correct Answer: B
Solution:
$ K=\frac{F}{x} $
$ =\frac{40}{2\times {{10}^{-2}}}=0.2\ N/m $ Work done$ =\frac{1}{2}Kx^{2}=\frac{1}{2}\times (0.2)\times {{(0.05)}^{2}}=2.5\ J $
BETA
