Respuesta :
Answer:
a) 6.25 m/s
b) f = 832 N
Explanation:
given data:
mass of child = 36 kg
length 2.94 m
tension = 416 N
For part (A), we need to add the forces in normal direction F(n),
[tex]F(n) = 2\times T - m\times g = m\times a[/tex]
where T is tension
m is mass, g = 9.81 m/s^2, and
a is acceleration is given as.
[tex]a = \frac{v^2}{r}[/tex]
where v is speed
r is the radius.
[tex]2\times T - m\times g = m\times \frac{v^2}{r} [/tex]
[tex]2*(416) - (36)\times (9.81) = (36)* \frac{v^2}{2.94}[/tex]
v = 6.25m/s
Part B)
F = 2*(416)
F = 832 N
Answer:v=6.25 m/s
Explanation:
Given
mass of child [tex]m=36 kg[/tex]
Length of chain [tex]L=2.94 m[/tex]
Tension in each chain [tex]T=416 N[/tex]
according to figure
[tex]2T-mg=\frac{mv^2}{L}[/tex]
[tex]\frac{mv^2}{L}=2T-mg[/tex]
[tex]\frac{36\times v^2}{2.94}=2\times 416-36\times 9.8[/tex]
[tex]v^2=\frac{479.2\times 2.94}{36}[/tex]
[tex]v=\sqrt{39.134}[/tex]
[tex]v=6.25 m/s[/tex]
Force exerted by the seat will be equal to the tension forces i.e. 2T
Force exerted by seat[tex]=2\times 416=832 N[/tex]
