Question

When a 13.8-kg block is attached to the free end of a vertically hanging spring, he...

When a 13.8-kg block is attached to the free end of a vertically hanging spring, he spring stretches by 21 cm. Wjat would the mass of the block have to be in order for the spring to stretch by 31.0 cm?

Homework Answers

Answer #1

Given
mass, m= 13.8 kg
x = 21 cm = 0.21 m
a) since F = kx
mg = kx
13.8 kg * 9.8 m/s^2 = k * 0.21 m
k = 644 N/m
b) period , T = 2π√m/k

   T = 2 * 3.14 *√13.8 kg/644 N/m

    T = 0.91 s

c)Frequency , f = 1/T = 1/0.91s

             f = 1.09 Hz

d)The amplitude is the maximum displacement from the equilibrium

position . so A = 31cm

e)Maximum speed , v = Aω = A(2π/T)

                                 = (0.31 m) (2 *3.14)/0.91s

                         v = 2.13 m/s

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