Respuesta :
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
