Answer:
a)v=5.81 m/s
b)N= 831.77 N
Explanation:
Given that
m = 39 kg
r= 2.98 m
T= 416 N
a)
Lets take speed of the boy at the lowest position is v m/s
The radial force Fc
[tex]F_c=\dfrac{mv^2}{r}[/tex]
The tension in the chain is T
[tex]2T-mg=\dfrac{mv^2}{r}[/tex]
Now by putting the values
[tex]2T-mg=\dfrac{mv^2}{r}[/tex]
[tex]2\times 416-39\times 10=\dfrac{39\times v^2}{2.98}[/tex]
v²=33.77
v=5.81 m/s
b)
Lets take normal force = N
[tex]N-mg=\dfrac{mv^2}{r}[/tex]
Now by putting the values
[tex]N-mg=\dfrac{mv^2}{r}[/tex]
[tex]N=mg +\dfrac{mv^2}{r}[/tex]
[tex]N=39\times 10+\dfrac{39\times 5.81^2}{2.98}[/tex]
N= 831.77 N
