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 = 0.25 m. While at this equilibrium position, the mass is then given an initial push downward at v = 5 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 = 0.43 s what is the speed of the block?
4)What is the magnitude of the maximum acceleration of the block?
5)At t = 0.43 s what is the magnitude of the net force on the block?

I found the solutions for 1 & 2.
1. -kx = ma
k = (6.3*-9.8)/-0.25 = 246.96 N/m

2. 1/2(pi) * sqrt(k/m) = 0.9963 Hz.

But I dont know how to find the solution for 3,4,5. Anyone can help me with this?
Thanks.

Explanation / Answer

The amplitude of the oscillation gives the total energy of the spring through the use of potential energy. So E = 1/2 A²k or A = sqrt(2E/k) = sqrt(mV²/k) = V sqrt(m/k) then use For speed, v(t) = -w*A*sin(w*t) For acceleration, amax = wˆ2*A For force, F(t) = k*A*sin(w*t) hope this helps please dont forget to rate my answer :)

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