Calculating Motor Input Power for a Lift

The problem asks us to find the input power required by a motor that is lifting a mass at a certain velocity with a given efficiency. We need to consider the work done against gravity and the efficiency of the motor.

Here's how we can solve this problem step-by-step:

• Identify the given values:
• Mass of the lift, $m = 200 \, kg$
• Velocity of the lift, $v = 5 \, m/s$
• Efficiency of the motor, $\eta = 80\% = 0.80$

Step 1: Calculate the Force Required

To lift the mass at a constant velocity, the motor must exert a force equal to the gravitational force acting on the mass. The gravitational force is given by $F = mg$, where $g$ is the acceleration due to gravity. We can take $g = 9.8 \, m/s^2$.

Force, $F = m \times g = 200 \, kg \times 9.8 \, m/s^2$

$F = 1960 \, N$

Step 2: Calculate the Output Power

The output power of the motor is the power delivered to the lift, which is used to lift the mass. Power is the rate at which work is done. When lifting a mass at a constant velocity, the power delivered is given by the product of the force and the velocity:

$P_{out} = Force \times Velocity = F \times v$

$P_{out} = 1960 \, N \times 5 \, m/s$

$P_{out} = 9800 \, W$

This is the power required to lift the mass, also known as the useful power output of the motor.

Step 3: Calculate the Input Power Using Efficiency

The efficiency of a motor is defined as the ratio of the output power to the input power. It is given by the formula:

$\eta = \frac{P_{out}}{P_{in}}$

We are given the efficiency ($\eta$) and we have calculated the output power ($P_{out}$). We can rearrange the formula to find the input power ($P_{in}$):

$P_{in} = \frac{P_{out}}{\eta}$

Substitute the values:

$P_{in} = \frac{9800 \, W}{0.80}$

$P_{in} = 12250 \, W$

Step 4: Convert Input Power to Kilowatts

The options are given in kilowatts (kW). To convert watts (W) to kilowatts (kW), we divide by 1000, since $1 \, kW = 1000 \, W$.

$P_{in} = \frac{12250}{1000} \, kW$

$P_{in} = 12.25 \, kW$

Therefore, the input power to the motor is 12.25 kW.