a motor lifts a 12 kg mass to a height of 15 meters at a content velocity in 3.5 seconds.
a) find the power
b) if the motor is only 35% efficient find the input energy
change in potential energy = m g h = 12 * 9.81 * 15 = 1766 Joules
power = joules per second = 1766/3.5 =505 Watts
Input power = 505/.15 = 3363 Watts
Input energy = 3363 Watts * 15 seconds = 50,451 Joules
To find the power of the motor, we can use the following formula:
Power = Work done / Time
First, let's find the work done by the motor to lift the mass. The work done is given by the formula:
Work = Force × Distance
In this case, the force can be calculated using Newton's second law of motion:
Force = Mass × Acceleration
Since the mass is 12 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we have:
Force = 12 kg × 9.8 m/s^2 = 117.6 N
Now, let's calculate the work done:
Work = Force × Distance = 117.6 N × 15 m = 1764 J (Joules)
To find the power, we divide the work done by the time taken:
Power = Work / Time = 1764 J / 3.5 s = 504 W (Watts)
So, the power output of the motor is 504 Watts.
b) To find the input energy, we need to consider the efficiency of the motor. Efficiency is defined as:
Efficiency = (Useful Output Energy / Total Input Energy) × 100
In this case, the useful output energy is the work done, which we calculated earlier as 1764 J. We need to find the total input energy.
Given that the motor is 35% efficient, we can rearrange the formula to solve for the input energy:
Total Input Energy = (Useful Output Energy / Efficiency) × 100
Total Input Energy = (1764 J / 0.35) × 100 = 5040 J
So, the input energy required by the motor is 5040 Joules.