Respuesta :
Answer:
a) Â T = 2.26 N, b) v = 1.68 m / s
Explanation:
We use Newton's second law
Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string
    sin 30 = [tex]\frac{T_x}{T}[/tex]
    cos 30 = [tex]\frac{T_y}{T}[/tex]
    Tₓ = T sin 30
    T_y = T cos 30
Y axis Â
    T_y -W = 0
    T cos 30 = mg           (1)
X axis
    Tₓ = m a
they relate it is centripetal
    a = v² / r
we substitute
     T sin 30 = m[tex]\frac{v^2}{r}[/tex]       (2)
a) we substitute in 1
     T = [tex]\frac{mg }{cos 30}[/tex]
     T = [tex]\frac{ 0.2 \ 9.8}{cos \ 30}[/tex]
     T = 2.26 N
b) from equation 2
      v² = [tex]\frac{T \ sin 30 \ r}{m}[/tex]
If we know the length of the string
     sin 30 = r / L
     r = L sin 30
we substitute
     v² = [tex]\frac{ T \ sin 30 \ L \ sin 30}{m}[/tex]
     v² = [tex]\frac{TL \ sin^2 30}{m}[/tex]
For the problem let us take L = 1 m
let's calculate
     v = [tex]\sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }[/tex]
     v = 1.68 m / s
