A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. The total mechanical energy is 25 J. The maximum speed of the block is:
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Ответ:
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
Ответ:
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