A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m. While at this equilibrium position, the mass is then given an initial push downward at v = 5.1 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 = 0.42 s what is the speed of the block?
4)What is the magnitude of the maximum acceleration of the block?
5)At t = 0.42 s what is the magnitude of the net force on the block?

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wagonbelleville wagonbelleville · Answer 1 · 2019-12-09T08:12:40+01:00

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

[tex]k = \dfrac{mg}{x}[/tex]

[tex]k = \dfrac{6.4 \times 9.8}{0.28}[/tex]

k = 224 N/m

b) [tex]f = \dfrac{\omega}{2\pi}[/tex]

[tex]\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s[/tex]

[tex]f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]

[tex]f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}[/tex]

[tex]f =0.94\ Hz[/tex]

c) [tex]v_b = -v cos \omega t[/tex]

[tex]v_b = -5.1 \times cos (5.92 \times 0.42)[/tex]

[tex]v_b = 4.04\ m/s[/tex]

d) [tex]a_{max} = v \omega[/tex]

[tex]a_{max} = 4.04 \times 5.92[/tex]

[tex]a_{max} =23.94\ m/s^2[/tex]

e) [tex]Y =- A sin (\omega t)[/tex]

[tex]A = \dfrac{v}{\omega}[/tex]

[tex]A = \dfrac{4.04}{5.92}[/tex]

A = 0.682 m

[tex]Y =- 0.682 \times sin (5.92 \times 0.42)[/tex]

[tex]Y =- 0.42[/tex]

Force =[tex]m \omega^2 |Y|[/tex]

=[tex]6.4 \times 5.92^2\times 0.42[/tex]

F = 94.20 N