A block with mass m =6.5 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =6.5 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.23 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.2 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 = 0.39 s what is the speed of the block?

4)

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

5)

At t = 0.39 s what is the magnitude of the net force on the block?

Explanation / Answer

a)

Spring Constant

K=mg/x =6.5*9.8/0.23

K=276.96 N/m

b)

angular frequency

W=sqrt[K/m]=sqrt[276.96/6.5]

W=6.53 rad/s

frequency

f=W/2pi=1.04 Hz

c)

V=VoCos(Wt) =4.2*Cos(6.53*0.39)

V=-3.48 m/s

d)

amax=VoW =4.2*6.53 =27.43m/s2

e)

a=-amaxSin(Wt)=-27.43*sin(6.53*0.39)

a=15.37 m/s2

F=ma =15.37*6.5

F=99.92 N

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