A lift arrangement comprises an electric motor, a cage and its counterweight - NSC Physical Sciences - Question 5 - 2017 - Paper 1

To calculate the work done by the gravitational force on the cage, we use the formula:

$W = F \cdot d \cdot \cos(\theta)$

where:

• $F$ is the weight of the cage and passengers, given by $F = mg = (1200\, \text{kg}) (9.8\, \text{m/s}^2) = 11760\, \text{N}$,
• $d$ is the distance moved upwards, which is 55 m,
• $heta$ is the angle between the force and direction of motion (180° in this case due to upward movement against gravity).

Thus, the work done is:

$W_{gravity} = 11760\, \text{N} \cdot 55\, \text{m} \cdot \cos(180°) = -646800\, ext{J}$

The work done by the counterweight is calculated similarly:

$W_{counterweight} = F \cdot d \cdot \cos(\theta)$

where the force from the counterweight is:

• $F = mg = (950\, \text{kg}) (9.8\, \text{m/s}^2) = 9100\, \text{N}$,
• $d$ is still 55 m, and
• $heta$ is 0° since the counterweight moves downwards in the same direction as the force.

Thus, the work done is:

$W_{counterweight} = 9100\, \text{N} \cdot 55\, \text{m} \cdot \cos(0°) = 500500\, ext{J}$

First, we determine the total work done, which is the work done against gravity minus the work done by the counterweight:

$W_{net} = W_{gravity} + W_{counterweight} = -646800\, ext{J} + 500500\, ext{J} = -146300\, ext{J}$

In this scenario, the total work that must be done by the motor takes into account the energy required to lift the cage:

$W_{motor} = -W_{net} = 146300\, ext{J}$

Next, we find the time taken in seconds:

• 3 minutes = 3 x 60 = 180 seconds.

Then, we can find the average power required using:

$P_{avg} = \frac{W_{motor}}{t} = \frac{146300\, ext{J}}{180\, ext{s}} \approx 812.78\, ext{W}$