Question

5. A 0.1 kg mass is attached to a spring which has a spring constant k = 20 N/mme initially stretched and released from rest. The mass continues to oscillate back and forth. a) If the spring is initially elongated 0.2m, what is the energy of the mass. The mass is released from rest. b) When the spring is elongated 0.15 m, what is the velocity of the mass? c) What is the maximum speed of the mass?

Answer 1

a)

emergy of the mass

U = 0.5 k x^2 = 0.5* 20* 0.2^2 = 0.4 J

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b)

velocity at that instant

v^2 = (k/m) (A^2 - x^2)

v^2 = ( 20 / 0.1)* ( 0.2^2 - 0.15^2)

v = 1.8708 m/s

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c)

Vo = A sqrt (k/m)

Vo = 0.2* sqrt ( 20/0.1)

Vo = 2.828 m/s

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