Two masses are connected by a light string passing over a frictionless pulley as shown on Figure. The M2= 6.1 kg mass is released from rest at height h = 2.8 m . We are given that M1=2.7 kg and I, the moment of inertia of the pulley is I = 0.198 kg m2 and its radius is R = 0.200 m.

a) What is in Joules, the initial gravitational potential energy of the system relative to the ground?

b) What is in Joules, the final (the instant m2 touches the ground) gravitational potential energy of the system relative to the ground?

c) What is in Joules, the final (the instant m2 touches the ground) kinetic energy of the system?

d) Determine the speed of M1 just as M2 hits the ground. Answer in units of m/s

To answer these questions, we need to apply the principles of conservation of energy. The total mechanical energy of a system remains constant if no external forces are acting on it. There are three forms of energy involved in this problem: gravitational potential energy, kinetic energy, and rotational kinetic energy.

a) To determine the initial gravitational potential energy of the system, we need to calculate the potential energy of both masses. The formula for gravitational potential energy is PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

For M1, the initial potential energy is PE1 = m1 * g * h = 2.7 kg * 9.8 m/s^2 * 2.8 m.

For M2, the initial potential energy is PE2 = m2 * g * h = 6.1 kg * 9.8 m/s^2 * 2.8 m.

The total initial potential energy of the system is the sum of both masses' potential energy: PE_initial = PE1 + PE2.

b) At the instant when M2 touches the ground, all of its gravitational potential energy is converted into kinetic energy. Therefore, the final gravitational potential energy of the system is zero.

c) The final kinetic energy of the system is the sum of the kinetic energies of both masses. The formula for kinetic energy is KE = 0.5 * m * v^2, where m is the mass and v is the velocity.

For M1, its final kinetic energy is KE1 = 0.5 * m1 * v1^2.

For M2, as it touches the ground, we can use the principle of conservation of mechanical energy to relate its potential energy at the top to its kinetic energy at the bottom. Since M2 is released from rest and there is no friction or energy loss, the potential energy at the top (PE2) equals the kinetic energy at the bottom (KE2). Therefore, the final kinetic energy of M2 is KE2 = PE2 = 0.

The total final kinetic energy of the system is the sum of both masses' kinetic energy: KE_final = KE1 + KE2.

d) To determine the speed of M1 just as M2 hits the ground, we can apply the principle of conservation of mechanical energy again. Since there is no external work or energy loss, the initial mechanical energy of the system (PE_initial) is equal to the final mechanical energy of the system, which is the sum of the final gravitational potential energy (zero) and the final kinetic energy (KE_final). Therefore, we have:

PE_initial = PE_final + KE_final.

Solving this equation will give us the value of v1, the velocity of M1 when M2 hits the ground.

Let's calculate these values:

a) PE1 = 2.7 kg * 9.8 m/s^2 * 2.8 m = 67.144 J
PE2 = 6.1 kg * 9.8 m/s^2 * 2.8 m = 163.912 J
PE_initial = PE1 + PE2 = 231.056 J

b) PE_final = 0 J

c) KE1 = 0.5 * 2.7 kg * v1^2
KE2 = 0 J
KE_final = KE1 + KE2 = KE1 = 0.5 * 2.7 kg * v1^2

d) PE_initial = PE_final + KE_final
231.056 J = 0 + 0.5 * 2.7 kg * v1^ 2
Solving for v1, we find:
v1 = sqrt((231.056 J) / (0.5 * 2.7 kg))

Note: Make sure to take the square root of the entire expression in the numerator before dividing by the denominator.

I hope this helps!