Two boxes are connected by by a massless string over a massless, frictionless pulley. Box 1 has a mass
of m1 = 15.0 kg and is initially at rest at the bottom of a frictionless surface inclined at an angle
θ = 30.0◦ above the horizontal. Box 2 has a mass of m2 = 10.0 kg and is initially at a height of 1.50
m above the bottom of the incline.
(a) What is the speed of box 2 when it hits the ground?
(b) What is the average power delivered to the system as box 2
To find the speed of box 2 when it hits the ground, we can use the principles of conservation of energy.
(a) First, let's determine the potential energy of box 2 when it is at a height of 1.50 m above the bottom of the incline. The potential energy is given by the formula:
Potential Energy = mass * acceleration due to gravity * height
Using the given values, the potential energy of box 2 is:
Potential Energy = 10.0 kg * 9.8 m/s^2 * 1.50 m = 147 J
Next, the potential energy of box 1 at the bottom of the incline is converted into kinetic energy of box 2 as it descends. The equation for kinetic energy is:
Kinetic Energy = 0.5 * mass * velocity^2
Since box 1 is initially at rest and there is no friction, all the potential energy is converted into kinetic energy.
Thus, we can equate the potential energy of box 2 to its kinetic energy at the bottom:
Potential Energy = Kinetic Energy
147 J = 0.5 * 10.0 kg * velocity^2
Simplifying the equation, we get:
velocity^2 = (2 * 147 J) / 10.0 kg
velocity^2 = 29.4 m^2/s^2
velocity = √(29.4 m^2/s^2)
velocity ≈ 5.42 m/s
Therefore, the speed of box 2 when it hits the ground is approximately 5.42 m/s.
(b) To find the average power delivered to the system as box 2 descends, we can use the work-energy principle.
The work done on the system is equal to the change in the system's mechanical energy.
The initial mechanical energy of the system is the potential energy of box 2 at the top:
Initial Mechanical Energy = 147 J
The final mechanical energy of the system is the sum of the kinetic energy of both boxes at the bottom:
Final Mechanical Energy = 0.5 * (15.0 kg + 10.0 kg) * (5.42 m/s)^2
Simplifying and calculating the final mechanical energy, we get:
Final Mechanical Energy = 410.92 J
The work done on the system is the difference between the initial and final mechanical energies:
Work Done = Final Mechanical Energy - Initial Mechanical Energy
Work Done = 410.92 J - 147 J
Work Done = 263.92 J
Finally, average power is given by the equation:
Average Power = Work Done / Time
Since the time is not provided, we cannot determine the average power without that information.
Please provide the time duration for which the system operates, and I can continue to calculate the average power delivered to the system as box 2 descends.