A block with mass m =6.9 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =6.9 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.27 m. While at this equilibrium position, the mass is then given an initial push downward at v = 3.9 m/s. The block oscillates on the spring without friction.

1.What is the spring constant of the spring?

2.What is the oscillation frequency?

3.After t = 0.35 s what is the speed of the block?

4.What is the magnitude of the maximum acceleration of the block?

5.At t = 0.35 s what is the magnitude of the net force on the block?

6.Where is the potential energy of the system the greatest?

a)At the highest point of the oscillation

b)At the new equilibrium position of the oscillation.

c)At the lowest point of the oscillation.

Please answer step by step thouroghly and I will rate 5 stars...plz help

Explanation / Answer

Ks=Mg

K=Mg/s

=6.9x9.8/0.27

=250.44

w=sqrt(k/m)

=6.02

the amplitude A can be calculated from energy conservation

1/2mv^2=1/2kA^2

A= 0.647

Velocity=Awsin(wt)

=0.647sin(6.02x0.35)

=1.98m/sec

a=Aw^2

=23.44

position after t=0.35 sec

x=Asinwt

=0.55

Net force= 139.4

potential energy is max at the top point of osscillation

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