Position #1:

radius, r₁ = 3 ft

Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is

ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:

Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively

I₁ = (4 lb)*(3 ft)² = 36 lb-ft²

I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,

I₁ω₁ = I₂ω₂

Therefore the angular velocity in position 2 is

ω₂ = (I₁/I₂)ω₁

= (36/16)*1 = 2.25 rad/s

The tangential velocity in position 2 is

v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.

In position 1, the centripetal force is

F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf

In position 2, the centripetal force is

F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.

Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is

W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer: 44.25 ft-lb