Answer:
The velocity is [tex]v_b = 20.17 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 0.245 \ kg[/tex]
The radius is [tex]r = 59.8 \ cm = 0.598 \ m[/tex]
The force is [tex]F = 30.9 \ N[/tex]
The speed of the ball is [tex]v = 16.0 \ m/s.[/tex]
Generally the kinetic energy at the top of the circle is mathematically represented as
[tex]K_t = \frac{1}{2} * m * v^2[/tex]
=> [tex]K_t = \frac{1}{2} * 0.245 * 16.0 ^2[/tex]
=> [tex]K_t = 31.36 \ J[/tex]
Generally the work done by the force applied on the ball from the top to the bottom is mathematically represented as
[tex]W = F * d[/tex]
Here d is the length of a semi - circular arc which is mathematically represented as
[tex]d = \pi * r[/tex]
So
[tex]W = 30.9 * 0.598 [/tex]
[tex]W = 18.48 \ J [/tex]
Generally the kinetic energy at the bottom is mathematically represented as
[tex]K_b = \frac{1}{2} * m * v_b^2[/tex]
=> [tex]K_b = \frac{1}{2} * 0.245 * v_b^2[/tex]
=> [tex]K_b = 0.1225 * v_b^2[/tex]
From the law of energy conservation
[tex]K_t + W =K_b[/tex]
=> [tex]31.36+ 18.48 = 0.1225 * v_b^2[/tex]
=> [tex]v_b = 20.17 \ m/s[/tex]
