A 4 kg block resting on a frictionless horizontal ledge is attached to a string that passes over a pulley and is attached to a hanging 2 kg block. The pulley is a uniform disk of radius 13.0 cm and mass 0.50 kg.

(a) Find the speed of the 2 kg block after it falls from rest a distance of 2.5 m.

(b) If an additional 5.0 mJ of energy are supplied to the rotational energy, what is the new angular speed of the ball?

look at the energy change:

change in PE=change in KE
2*g*2.5=1/2 (4+2)v^2 + 1/2 I w^2

where I is the moment of inertia for a pulley ( I would model it as a solid disk)
now to relate w to v, remember w=v/r

b) makes no sense to me.

To solve this problem, we need to apply the principle of conservation of mechanical energy, as well as the principle of conservation of angular momentum.

(a) First, let's find the initial potential energy of the 2 kg block:

Potential energy = mass * gravitational acceleration * height
Initial potential energy = 2 kg * 9.8 m/s^2 * 2.5 m
Initial potential energy = 49 J

The initial potential energy will be converted into the kinetic energy of the 2 kg block after it falls. Since there is no friction, the total mechanical energy will be conserved.

Total mechanical energy = kinetic energy of the 4 kg block + kinetic energy of the pulley + kinetic energy of the 2 kg block

The kinetic energy of the 4 kg block can be calculated using the formula:

Kinetic energy = (1/2) * mass * velocity^2

However, since the 4 kg block is at rest, its initial kinetic energy is zero.

The kinetic energy of the pulley can be calculated using the formula:

Kinetic energy = (1/2) * moment of inertia * angular velocity^2

The moment of inertia of a uniform disk is given by the formula:

Moment of inertia = (1/2) * mass * radius^2

Substituting the given values:

Moment of inertia = (1/2) * 0.50 kg * (0.13 m)^2
Moment of inertia = 0.008125 kg⋅m^2

Knowing that the velocity of the 4 kg block is equal to the velocity of the pulley's edge, we can determine that the kinetic energy of the pulley is equal to the change in potential energy (initial potential energy) of the 2 kg block:

(1/2) * moment of inertia * angular velocity^2 = initial potential energy
(1/2) * 0.008125 kg⋅m^2 * angular velocity^2 = 49 J

Now, let's solve for the angular velocity:

angular velocity^2 = (2 * initial potential energy) / (moment of inertia)
angular velocity^2 = (2 * 49 J) / (0.008125 kg⋅m^2)
angular velocity^2 = 12000 rad/s

Taking the square root:

angular velocity = sqrt(12000 rad/s)
angular velocity ≈ 109.54 rad/s

Since the angular velocity of the pulley and the linear velocity of the 4 kg block are the same, we can now find the linear velocity of the 2 kg block by using the formula:

velocity = angular velocity * radius
velocity = 109.54 rad/s * 0.13 m
velocity ≈ 14.24 m/s

Therefore, the speed of the 2 kg block after falling a distance of 2.5 m is approximately 14.24 m/s.

(b) To find the new angular speed of the ball after supplying an additional 5.0 mJ of energy to the rotational energy, we need to calculate the new total energy and use the principle of conservation of energy.

The additional energy added is given as 5.0 mJ, which is equivalent to 0.005 J.

The new total mechanical energy is the sum of the initial mechanical energy and the additional energy:

New total mechanical energy = initial mechanical energy + additional energy
New total mechanical energy = 49 J + 0.005 J
New total mechanical energy = 49.005 J

Since the total mechanical energy is conserved, we can write:

Total mechanical energy = kinetic energy of the 4 kg block + kinetic energy of the pulley + kinetic energy of the 2 kg block

Using the formulas for the kinetic energy of the 4 kg block and the kinetic energy of the pulley, we can rewrite the equation:

Total mechanical energy = (1/2) * mass * velocity^2 + (1/2) * moment of inertia * angular velocity^2 + kinetic energy of the 2 kg block

The kinetic energy of the 4 kg block and the kinetic energy of the pulley remain the same, so the only term that changes is the kinetic energy of the 2 kg block.

Let's denote the new angular velocity as ω_new. The new kinetic energy of the 2 kg block can be calculated as:

New kinetic energy of the 2 kg block = (1/2) * mass * velocity^2 = (1/2) * 2 kg * ω_new^2

Now, let's substitute the known values and solve for the new angular velocity:

New total mechanical energy = (1/2) * mass * velocity^2 + (1/2) * moment of inertia * ω_new^2 + (1/2) * 2 kg * ω_new^2
49.005 J = 0 + (1/2) * 0.008125 kg⋅m^2 * ω_new^2 + (1/2) * 2 kg * ω_new^2

Simplifying the equation:

49.005 J = 0.0040625 kg⋅m^2 * ω_new^2 + 1 kg * ω_new^2
49.005 J = 0.0040625 kg⋅m^2 * ω_new^2 + 1000 kg⋅m^2/s^2 * ω_new^2
49.005 J = 1000.0040625 kg⋅m^2/s^2 * ω_new^2

Solving for ω_new^2:

ω_new^2 = (49.005 J) / (1000.0040625 kg⋅m^2/s^2)
ω_new^2 ≈ 0.049 J / 1 kg⋅m^2/s^2
ω_new^2 ≈ 0.049 s^2

Taking the square root:

ω_new ≈ sqrt(0.049 s^2)
ω_new ≈ 0.221 s^(-1) rad/s

Therefore, the new angular speed of the ball after supplying an additional 5.0 mJ of energy is approximately 0.221 rad/s.

To solve this problem, we need to apply the principles of conservation of energy and rotational motion.

(a) To find the speed of the 2 kg block after it falls a distance of 2.5 m, we can use the conservation of mechanical energy. The initial mechanical energy is equal to the final mechanical energy. Initially, the system has gravitational potential energy due to height and no kinetic energy. Finally, the 2 kg block will have both kinetic energy and gravitational potential energy.

The gravitational potential energy of the 4 kg block initially is given by:
PE1 = m1 * g * h1
where m1 is the mass of the 4 kg block, g is the acceleration due to gravity, and h1 is the initial height of the 4 kg block. As it is resting on the ledge, h1 is zero, so the gravitational potential energy initially is zero.

The gravitational potential energy of the 2 kg block finally is given by:
PE2 = m2 * g * h2
where m2 is the mass of the 2 kg block, g is the acceleration due to gravity, and h2 is the final height of the 2 kg block. The final height is given as 2.5 m.

Since the system is frictionless, the energy lost due to friction is negligible. Therefore, the initial mechanical energy is equal to the final mechanical energy.

Initially, the system has zero kinetic energy and zero gravitational potential energy. Finally, the system will have kinetic energy due to the motion of the 2 kg block and gravitational potential energy due to the height of the 2 kg block.

The kinetic energy of the 2 kg block finally is given by:
KE = (1/2) * m2 * v^2
where v is the final velocity of the 2 kg block.

Using the conservation of energy, we can write:
PE1 + KE1 = PE2 + KE2

Since the 4 kg block starts from rest, its initial kinetic energy is zero. Therefore, the equation becomes:
0 + 0 = m2 * g * h2 + (1/2) * m2 * v^2

Simplifying the equation, we can solve for v:
v = sqrt(2 * g * h2)
where sqrt denotes square root, g is the acceleration due to gravity, and h2 is the final height of the 2 kg block (2.5 m).

Let's calculate the value of v using the above formula:
v = sqrt(2 * 9.8 m/s^2 * 2.5 m)
v = sqrt(49 m^2/s^2)
v ≈ 7 m/s

Therefore, the speed of the 2 kg block after it falls a distance of 2.5 m is approximately 7 m/s.

(b) To find the new angular speed of the ball when an additional 5.0 mJ of energy is supplied to the rotational energy, we need to use the principles of rotational motion.

The rotational kinetic energy of a rotating object is given by:
KE_rot = (1/2) * I * ω^2
where KE_rot is the rotational kinetic energy, I is the moment of inertia, ω is the angular speed.

Since the pulley is a uniform disk, the moment of inertia can be calculated as:
I = (1/2) * m * r^2
where m is the mass of the pulley disk and r is the radius of the pulley disk.

Given that the additional 5.0 mJ of energy is supplied to the rotational energy, we can write:
KE_rot_new = KE_rot_initial + additional_energy

Substituting the corresponding values:
(1/2) * I * ω^2_new = (1/2) * I * ω^2_initial + additional_energy

Simplifying the equation, we can solve for ω^2_new:
ω^2_new = ω^2_initial + (2 * additional_energy) / I

The angular speed ω is given by:
ω = v / r
where v is the linear velocity and r is the radius of the pulley disk.

Substituting the value of ω into the equation for ω^2_new, we get:
ω^2_new = (v / r)^2 + (2 * additional_energy) / I

Let's calculate the value of ω^2_new using the above formula:
ω^2_new = (7 m/s / 0.13 m)^2 + (2 * 5.0 * 10^-3 J) / [(1/2) * 0.50 kg * (0.13 m)^2]
ω^2_new = (3780 m^2/s^2) + (2 * 5.0 * 10^-3 J) / [0.1625 kg m^2]
ω^2_new ≈ 3780 m^2/s^2 + 0.0615 m^2/s^2

Therefore, the new angular speed of the ball when an additional 5.0 mJ of energy is supplied to the rotational energy is approximately 3780.0615 m^2/s^2.