Step 1

We are given the weight of the ball $W = 4\;{\rm{lb}}$, the radius of the circle as ${r_1} = 3\;{\rm{ft}}$, the initial speed of the ball as ${\left( {{v_B}} \right)_1} = 6\;{\rm{ft/s}}$, and the velocity of the cord as ${v_r} = 2\;{\rm{ft/s}}$.

We are asked to determine the speed of the ballwhen the radius is ${r_2} = 2\;{\rm{ft}}$ and the work done to pull down the cord.

 
Step 2

We will apply the conservation of angular momentum to find the transverse component of the final speed of the ball.

\[\begin{array}{c} {H_1} = {H_2}\\ m{\left( {{v_B}} \right)_1}{r_1} = m{\left( {{v_\theta }} \right)_2}{r_2}\\ {\left( {{v_B}} \right)_1}{r_1} = {\left( {{v_\theta }} \right)_2}{r_2} \end{array}\] … (1)

Here, ${H_2}$ is the final angular momentum, ${H_1}$ is the initial angular momentum, $m$ is the mass of the balland ${\left( {{v_\theta }} \right)_2}$ is the transverse component of the final speed of the particle.

Substitute the given value in equation (1) to find the final time required by the car.

\[\begin{array}{c} \left( {6\;{\rm{ft/s}}} \right) \times \left( {3\;{\rm{ft}}} \right) = {\left( {{v_\theta }} \right)_2} \times \left( {2\;{\rm{ft}}} \right)\\ {\left( {{v_\theta }} \right)_2} = 9\;{\rm{ft/s}} \end{array}\]
 
Step 3

We will find the final speed of the ball at point at ${r_2} = 2\;{\rm{ft}}$.

\[{v_2} = \sqrt {\left( {{v_r}} \right)_2^2 + \left( {{v_\theta }} \right)_2^2} \]

Here, ${\left( {{v_r}} \right)_2}$ is the radial component of the final speed of the particle.

Substitute the given value in the above equation.

\[\begin{array}{c} {v_2} = \sqrt {{{\left( {2\;{\rm{ft/s}}} \right)}^2} + {{\left( {9\;{\rm{ft/s}}} \right)}^2}} \\ {v_2} = 9.22\;{\rm{ft/s}} \end{array}\]
 
Step 4

We will apply the principal of work and energy to find the work done required to pull down the cord.

\[\begin{array}{c} {T_1} + \sum {{U_{1 - 2}}} = {T_2}\\ \frac{1}{2}mv_1^2 + \sum {{U_{1 - 2}}} = \frac{1}{2}mv_2^2\\ \frac{1}{2}\left( {\frac{W}{g}} \right)v_1^2 + \sum {{U_{1 - 2}}} = \frac{1}{2}\left( {\frac{W}{g}} \right)v_2^2\\ \sum {{U_{1 - 2}}} = \left( {\frac{W}{{2g}}} \right)\left( {v_2^2 - v_1^2} \right) \end{array}\] …. (2)

Here, ${T_1}$ is the initial kinetic energy, ${T_2}$ is the final kinetic energy and $\sum {{U_{1 - 2}}} $ is the work done to pull down the cord.

Substitute the given value in equation (2) descend.

\[\begin{array}{c} \sum {{U_{1 - 2}}} = \left( {\frac{{{\rm{4}}\;{\rm{lb}}}}{{2 \times 32.2\;{\rm{ft/}}{{\rm{s}}^{\rm{2}}}}}} \right)\left( {{{\left( {9.22\;{\rm{ft/s}}} \right)}^2} - {{\left( {6\;{\rm{ft/s}}} \right)}^2}} \right)\\ \sum {{U_{1 - 2}}} = 3.04\;{\rm{lb}} \cdot {\rm{ft}} \end{array}\]