A block with mass m =6.2 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =6.2 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.22 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.6 m/s. The block oscillates on the spring without friction.

What is the spring constant of the spring? 276.52 N/m

What is the oscillation frequency? 1.063? Hz

a.) After t = 0.33 s what is the speed of the block?

b.) What is the magnitude of the maximum acceleration of the block?

c.) At t = 0.33 s what is the magnitude of the net force on the block?

d.) Where is the potential energy of the system the greatest?

Explanation / Answer

A) angular frequency w = 2pi*f = 6.28*1.063 = 6.679 rad/s

from given data V= wA

A = 4.6/6.679 = 0.689 m

from x = A sinwt

v = wA cos(wt)

at t =0.33 sec

v = (6.679)0.689*cos(6.679*0.33) = 2.72 m/s

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B)

max acceleration, a = -w^2*A = 30.735 m/s^2

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C)

at t=0.33 sec

position x = 0.22 m

acceleration a = -w^2*x = 9.81 m/s^2

force F = ma = 6.2*(9.81) = 60.84 N

D)

potential energy =0.5*kA^2 = 0.5*276.52*0.689^2 = 65.63 J

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