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

The work done by the gravitational force on an object can be calculated using the formula:

$W = F imes d imes ext{cos}( heta)$

Where:

• $W$ is the work done
• $F$ is the force applied
• $d$ is the distance moved
• $heta$ is the angle between the force and the direction of motion

For the cage:

• The gravitational force ($F_{gravity}$) acting on the cage is: $F_{gravity} = m_{ ext{cage}} imes g = 1200 ext{ kg} imes 9.8 ext{ m/s}^2 = 11760 ext{ N}$

Calculating the work done on the cage moving upwards 55 m: $W_{gravity} = (1200)(9.8)(55) ext{ Joules} = 646800 ext{ J}$

$W_{gravity} = 646800 ext{ J}$

For the counterweight:

• The gravitational force ($F_{counterweight}$) is: $F_{counterweight} = m_{ ext{counterweight}} imes g = 950 ext{ kg} imes 9.8 ext{ m/s}^2 = 9310 ext{ N}$

Calculating the work done by the counterweight moving down 55 m: $W_{counterweight} = (950)(9.8)(55) ext{ Joules} = 512050 ext{ J}$

The total work done during the operation of the lift system is the sum of the work done on both the cage and the counterweight:

$W_{total} = W_{gravity} - W_{counterweight} = 646800 ext{ J} - 512050 ext{ J} = 134750 ext{ J}$

To find the average power ( ext{P}_{ave}), divide the total work done by the time taken (in seconds): ext{P}_{ave} = rac{W_{total}}{ ext{t}} = rac{134750}{180} ext{ W} \\ ext{P}_{ave} ext{ = 748.61 W}