Find how much work ∆<em>W</em> is done by the motor in lifting the elevator:
<em>P</em> = ∆<em>W</em> / ∆<em>t</em>
where
• <em>P</em> = 45.0 kW = power provided by the motor
• ∆<em>W</em> = work done
• ∆<em>t</em> = 20.0 s = duration of time
Solve for ∆<em>W</em> :
∆<em>W</em> = <em>P</em> ∆<em>t</em> = (45.0 kW) (20.0 s) = 900 kJ
In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is <em>M</em> = (<em>m</em> + 490.0) kg, where <em>m</em> is the mass of the elevator alone. Then
∆<em>W</em> = <em>M</em> <em>g h</em>
where
• <em>g</em> = 9.80 m/s² = acceleration due to gravity
• <em>h</em> = 35.0 m = distance covered by the elevator
Solve for <em>M</em>, then for <em>m</em> :
<em>M</em> = ∆<em>W</em> / (<em>g h</em>) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg
<em>m</em> = <em>M</em> - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg
