A 280 electric motor is 90% efficient, meaning that 90% of the energy supplied to it ends up as mechanical work.
If the motor lifts a 210 weight at 3.5 , how much current does it draw?
Express your answer using two significant figures.
I= ? Amps
No units are given; therefore, I'll have to make some assumptions:
Wt = 210 Newtons @ 3.5m/s.
280 Volt motor
Power = 210 * 3.5m/s = 735 Watts output
Power Input = Po / 0.9 = 735/0.9 = 817
Watts.
P = V*I,
I = P / V = 817 / 280 = 2.9 Amps.
To find the current (I) drawn by the motor, we can use the following formula:
Power (P) = Voltage (V) * Current (I)
In this case, the power input to the motor is given by:
Power (P) = 280 W
Since the motor is 90% efficient, we can calculate the power output as follows:
Power output = Efficiency * Power input
= 0.90 * 280 W
= 252 W
Now, we can rearrange the formula to solve for current (I):
I = Power (P) / Voltage (V)
We need to determine the voltage applied to the motor. The weight lifted by the motor indicates that it is doing work against gravity. The formula for work (W) is given as:
Work = Force (F) * Distance (d)
In this case, the force is the weight and the distance is the height lifted. Therefore,
Work = Force (F) * Distance (d)
= 210 N * 3.5 m
= 735 J (Joules)
Since work is equal to the power output (in this scenario), we can equate the expressions:
Power output = Work
= 735 W
Now, we can substitute the values into the formula for current (I):
I = Power (P) / Voltage (V)
I = 252 W / V
From the given information, we do not have the voltage provided. Hence, we cannot determine the current drawn by the motor without knowing the voltage.