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3 years ago

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A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. Th

e total mechanical energy is 25 J. The maximum speed of the block is:

1 answer:

wolverine [178]3 years ago

6 0

Answer:

Explanation:

easy way

when system is all kinetic energy, velocity is at a maximum

E = ½mv²

v = √(2E/m) = √(2(25)/0.5) = √100 = 10 m/s

harder way

ω = √(k/m) = √(80/0.5) = √160 rad/s

When the system is entirely spring potential, the amplitude A is

E = ½kA²

A = √(2E/k) = √(2(25)/80) = 0.790569... = 0.79 m

maximum velocity is ωΑ = 0.79√160 = 10 m/s

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The question is incomplete. Here is the complete question.

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