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A spring of negligible mass stretches 2.20 cm from its relaxed length when a force of 10.5 N is applied. A 1.1 kg particle rest...Asked by Anonymous
A spring of negligible mass stretches 2.20 cm from its relaxed length when a force of 10.50 N is applied. A 1.100 kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x=5.0 cm and released from rest at t=0.
(c) What is the total energy of the system?
(c) What is the total energy of the system?
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Answered by
drwls
Total energy = Initial potential energy
= (1/2)kX^2
k = 10.50/0.22 = 47.7 N/m
X = initial displacement = 0.05 m
= (1/2)kX^2
k = 10.50/0.22 = 47.7 N/m
X = initial displacement = 0.05 m
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