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A spring with k=33.5 N/m has a 1.20 kg mass attached. it is pulled 0.120 m and released. What is its maximum speed

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Answer: v = 0.634 m/s

Step-by-step explanation:

The maximum speed of the mass will be obtained when the system passes through equilibrium and thus there is no elastic potential energy. In the beginning, there is no kinetic energy but only potential energy. The total energy in the system at these two times is equal:

0.5kx^2 = 0.5mv^2

kx^2 = mv^2

v = sqrt(kx^2/m)

v = sqrt((33.5)(0.120)^2/(1.20))

v = 0.634 m/s

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