A block with mass m =6.3 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =6.3 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.24 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.8 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?

Explanation / Answer

1) k = mg/x = (6.3*9.8)/0.24 = 257.25 N/m

2) f = (1/2pi)*sqrt(k/m) = 1.017 Hz

3) Think of it this way you know that the velocity is 4.8 m/s so use that as your max v for now which in following equation equals A so the equation is
V= -A sin(t + ).
so we know that must equal pi/2 so with that said

V= -4.8sin(6.39*0.35 + pi/2)
=-4.8sin(3.807)
=-4.2*-0.6176127
=2.6 m/s

4) from the equation A= -^2A cos(t + )
^2A equals the max a so you need to just find that
we know = 6.39 and we know that v=A so we can say A=v/
^2A=> ^2(v/)=>v=> 16.614 m/s2

5) F = ma = 6.3*16.614 = 104.6682 N

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