Answer:
The value is [tex]u = 27.93 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 0.250 \ kg[/tex]
The radius of the circle is [tex]r = 0.5 \ m[/tex]
The force exerted is [tex]F = 33 \ N[/tex]
The speed of the ball on top of the circle is [tex]v = 13.0 \ m/s[/tex]
Gnerally from the law of energy conservation we have that
[tex]PE + W = \Delta KE[/tex]
Here PE is the potential energy of the pitcher which is mathematically represented as
[tex]PE = m * g * (2 r )[/tex]
=> [tex]PE = 0.250 * 9.8 * (2 * 0.5 )[/tex]
=> [tex]PE = 24.5\ J[/tex]
And W is the workdone by the force applied which is mathematically represented as
[tex]W = F * \pi * r[/tex]
=> [tex]W = 33 * 3.142 * 0.5[/tex]
=> [tex]W = 51.84 \ J[/tex]
And [tex]\Delta KE[/tex] is the change in kinetic energy which is mathematically represented as
[tex]\Delta KE = \frac{1}{2} * m * (v^2 - u^2 )[/tex]
=> [tex]\Delta KE = \frac{1}{2} * 0.250 * ( 13^2 - u^2 )[/tex]
=> [tex]\Delta KE = 0.125 * ( 13^2 - u^2 )[/tex]
=> [tex]\Delta KE = 21.125 - 0.125u^2[/tex]
So
[tex]24.5 + 51.84 = 21.125 - 0.125u^2[/tex]
=> [tex]u = 27.93 \ m/s[/tex]
The velocity of the ball released at the bottom is required.
The velocity of the ball released at the bottom is 24.56 m/s
F = Force = 33 N
r = Radius of loop = 0.5 m
m = Mass of ball = 0.25 kg
u = Initial velocity of ball = 13 m/s
v = Final velocity
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
The energy balance of the system is given by
[tex]F\pi r+mg2r=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow v=\sqrt{\dfrac{2}{m}(F\pi r+mg2r)+u^2}\\\Rightarrow v=\sqrt{\dfrac{2}{0.25}(33\times \pi\times 0.5+0.25\times 9.81\times 2\times 0.5)+13^2}\\\Rightarrow v=24.56\ \text{m/s}[/tex]
The velocity of the ball released at the bottom is 24.56 m/s.
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