A loaded ore car has a mass of 935 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 26.0° above the horizontal. The car accelerates uniformly to a speed of 2.20 m/s in 11.0 s and then continues at constant speed
(b) What maximum power must the winch motor provide?
(c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length 1100 m?
I figured out in part A the power when the car is at constant speed but I can't figure out these last parts.
To solve parts (b) and (c), we will need to use the concepts of work, power, and energy.
For part (b), we need to find the maximum power the winch motor must provide. Power is defined as the rate at which work is done or the rate at which energy is transferred. In this case, while the car is accelerating, the winch motor is doing work to overcome the gravitational force acting on the car.
To find the maximum power, we can use the formula for power:
Power = Work / Time
The work done by the winch motor is equal to the change in kinetic energy of the car during the acceleration phase. Since the car starts from rest and reaches a speed of 2.20 m/s, the change in kinetic energy can be calculated using the formula:
Change in Kinetic Energy = (1/2) * Mass * (Final Velocity)^2
Plugging in the given values, we get:
Change in Kinetic Energy = (1/2) * 935 kg * (2.20 m/s)^2
Next, we divide the change in kinetic energy by the time it takes for the car to accelerate (11.0 s) to find the power during the acceleration phase.
Power = Change in Kinetic Energy / Time
Now, let's calculate the maximum power required by the winch motor using the given values:
Power = [(1/2) * 935 kg * (2.20 m/s)^2] / 11.0 s
Simplifying the equation, we get the answer to part (b) which is the maximum power the winch motor must provide.
For part (c), we need to determine the total energy transferred out of the motor by work by the time the car moves off the end of the track. Since the car is at constant speed after the acceleration phase, its kinetic energy remains constant.
The total energy transferred out of the motor by work is equal to the work done against gravity to raise the car to the final height. This can be calculated using the formula:
Work = Force * Distance
The force acting on the car is the component of the gravitational force parallel to the incline, which can be determined as:
Force = Mass * g * sin(theta)
where g is the acceleration due to gravity and theta is the angle of inclination (26.0°).
The distance traveled by the car can be calculated by using the given length of the track (1100 m).
Now, we can calculate the work done against gravity using the formula:
Work = (Mass * g * sin(theta)) * Distance
Substituting the given values, we get the answer to part (c), which is the total energy transferred out of the motor by work by the time the car moves off the end of the track.
Remember to use the appropriate units in all calculations.