A puck (mass m1 = 2.20 kg) slides on a frictionless table as shown in the figure below. The puck is tied to a string that runs through a hole in the table and is attached to a mass m2 = 6.6 kg. The mass m2 is initially at a height of h = 6.6 m above the floor with the puck traveling in a circle of radius r = 1.76 m with a speed of 6.6 m/s. The force of gravity then causes mass m2 to move downward a distance 0.66 m.

angular momentum is conserved since there is no torque in this problem

I omega = constant
m R^2 omega = constant
m R v = constant
m is constant
so
R v = constant

original R = 1.76
final R = 1.76 - .66 = 1.1

so what ever your question might be (you did not ask)

1.76 (6.6) = 1.1 (v)
where v is the final tangential velocity

To find the work done by gravity on mass m2 as it moves downward, we can use the formula:

W = m2 * g * d

where:
W = work done by gravity
m2 = mass of the object (6.6 kg)
g = acceleration due to gravity (-9.8 m/s^2)
d = distance moved downward (0.66 m)

Plugging in the given values, we get:

W = 6.6 kg * (-9.8 m/s^2) * 0.66 m

W = -42.47 Joules

Therefore, the work done by gravity on mass m2 as it moves downward is -42.47 Joules.

To determine the final speed of the puck and the tension in the string after the mass m2 moves downward by a distance of 0.66 m, we can use the principle of conservation of mechanical energy.

First, let's analyze the initial state of the system. The puck is moving in a circle of radius 1.76 m with a speed of 6.6 m/s. Since the table is frictionless, the only forces acting on the puck are the tension in the string and the gravitational force.

At the initial state, the gravitational potential energy of mass m2 is given by:
PE_initial = m2 * g * h

Where m2 is the mass of mass m2, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the initial height (6.6 m). We can calculate this as:
PE_initial = 6.6 kg * 9.8 m/s^2 * 6.6 m

Next, let's consider the final state of the system after mass m2 moves downward by a distance of 0.66 m. At this position, the tension in the string accelerates mass m1 along the circle.

The final gravitational potential energy of mass m2 is given by:
PE_final = m2 * g * (h - 0.66 m)

The change in gravitational potential energy is given by:
ΔPE = PE_final - PE_initial

Since mechanical energy is conserved, the work done by tension in the string is equal to the change in gravitational potential energy:
Work = ΔPE = T * 0.66 m

Where T is the tension in the string.

We also know that the work done by the tension is equal to the change in kinetic energy of the puck:
Work = ΔKE = (1/2) * m1 * v^2 - (1/2) * m1 * u^2

Where m1 is the mass of the puck, v is the final speed of the puck, and u is the initial speed of the puck as it moves along the circle.

Since the puck is moving in a circle, the initial speed u is given by:
u = r * ω

Where r is the radius of the circle and ω is the angular speed of the puck.

To find ω, we can use the relationship between linear speed v and angular speed ω:
v = r * ω

Now, we have two equations:
Work = (1/2) * m1 * v^2 - (1/2) * m1 * u^2
Work = T * 0.66 m

We can solve these equations simultaneously to find the final speed v and the tension in the string T.