Answer:
a
[tex]v_r =8.65 \ ft/s [/tex]
b
[tex] W_{1-2} = 3.24 \ ft \cdot lb[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 4 \ lb[/tex]
The radius is [tex]r= 3 \ ft[/tex]
The speed is [tex]v_B_1 = 4.8 \ ft /s[/tex]
The speed of the attached cord is [tex]v_c =2.2 \ ft[/tex]
The position that is been considered is [tex]r_1 = 2 \ ft[/tex]h
Generally according to the law of angular momentum conservation
[tex]L_a = L_b[/tex]
Here [tex]L_a[/tex] is the initial momentum of the ball which is mathematically represented as
[tex]L_a = m* v_B_1 * r[/tex]
while
[tex]L_b[/tex] is the momentum of the ball at r = 2 ft which is mathematically represented as
[tex]L_a = m* v_B_2 * r_1[/tex]
So
[tex]m* v_B_1 * r = m* v_B_2 * r_1[/tex]
=> [tex] 4.8 * 3 = v_B_2 * 2[/tex]
=> [tex] v_B_2 = 7.2 \ ft/s [/tex]
Generally the resultant velocity of the ball is
[tex]v_r = \sqrt{v_B_2^2 + v_B_1^2 }[/tex]
=> [tex]v_r = \sqrt{7.2^2 + 4.8^2 }[/tex]
=> [tex]v_r =8.65 \ ft/s [/tex]
Generally according to equation for principle of work and energy we have that
[tex]K_1 + \sum W_{1-2} = K_2[/tex]
Here [tex]K_1[/tex] is the initial kinetic energy of the ball which is mathematically represented as
[tex]K_1 = \frac{1}{2} * m* v_B_1^2[/tex]
While [tex]\sum W_{1-2}[/tex] is the sum of the total workdone by the ball
and [tex]K_2[/tex] is the final kinetic energy of the ball which is mathematically represented as [tex]K_2 = \frac{1}{2} * m* v_r^2[/tex]
So
[tex]\sum W_{1-2} = \frac{1}{2} * m (v_r^2 - v_B_1^2)[/tex]
Here m is the mass which is mathematically represented as
[tex]m = \frac{W}{g}[/tex] here W is the weight in lb and g is the acceleration due to gravity which is [tex]g = 32 \ ft/s^2[/tex]
So
[tex]\sum W_{1-2} = \frac{1}{2} * \frac{4}{32} * (8.65^2 - 4.8^2)[/tex]
=> [tex] W_{1-2} = 3.24 \ ft \cdot lb[/tex]
