An object of mass 0.60 kg hangs from the end of a spring. Imagine that the objec

Question

An object of mass 0.60 kg hangs from the end of a spring. Imagine that the object is lifted upward and held at rest at the position where the string is not stretched. then the object is released. It is observed that the lowest point in the object's subsequent oscillation is at .120 m below the point of where it was released. A) What is the amplitude of the oscillation? B What is the spring constant at the spring? C)What is the object's maximum speed? Please explain your reasoning and show your work for each step

Explanation / Answer

B)by applying conservation of energy

we have

0.5k*A2= mgA where A is the amplitude and hence

k=mg/0.5A=98N/m

a) mg=kx==> x=6cm is where the block rests that is point about which it oscillates hence

amplitude 12-6=6cm

c)0.5mv2max = mg*A-0.5*kA2 ,A=0.06m

hence on solving we get vmax= 0.77m/s

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