A force of 60 N stretches a spring 30cm. A mass of 4 kg is hung from the spring and comes to rest.

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A force of 60 N stretches a spring 30cm. A mass of 4 kg is hung from the spring and comes to rest.

[From: ] [author: ] [Date: 12-01-09] [Hit: ]

889 + 0.212 = 1.1 seconds.Part f tells me that part e is the two stay together. So I will only put thought into part f.The reason I did this entire thing as a differential equation was this part,......

The tension is just a force so F = ma = (4kg)(-2.5 m/s/s) = -10N

For part d we must be fully aware that the first instance of getting towards 5cm above equilibrium will occur before the spring ever hit equilibrium. This means we must take our 0.212043 s that we've been working with and add a period to it.

So part d the answer is 0.889 + 0.212 = 1.1 seconds.

Part f tells me that part e is the two stay together. So I will only put thought into part f.

The reason I did this entire thing as a differential equation was this part, actually. We know the force at any given time will be represented by:

F(t) = ma
F(t) = m(x''(t))
F(t) = -157.5cos(7.07 t)

So if we doubled the amplitude our force function would change to:

F(t) = -315cos(7.07 t)

An object of mass < 4.5 kg would not be able to stay on simply due to the fact the force tops out at 32gs!

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what does c represent?

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C is the dampening coefficient. There was none in this question so I just dropped it out.

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ok i understand

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for part f is it referring to the time or distance from equilibrium that it woul begin to seperate?

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Part f was from the beginning when it was 10cm from equilibrium. If you want it to start from equilibrium just add 0.222144 inside the cosine function to do a phase shift.

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Part f was from the beginning when it was 10cm from equilibrium. If you want it to start from equilibrium just add 0.222144 inside the cosine function to do a phase shift.

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