What is the potential energy of a spring that is stretched 0.15 m from equilibrium and has a spring constant of 0.55 N/m?
1 answer:
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Answer:
390 J
Explanation:
m = 3 kg
u = 16 i + 2 j
(a) Magnitude of velocity = = 16.1245 m/s
KEi = 1/2 m v^2 = 0.5 x 3 x 16.1245 = 390 J
(b) v = 18 i + 14 j
Magnitude of velocity = = 22.804 m/s
KEf = 1/2 m v^2 = 0.5 x 3 x 22.804 = 780 J
According to the work energy theorem
Work done = change in KE = KEf - KEi = 780 - 390 = 390 J