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Answer :
The work done by the cable on a 1,438-kg elevator to lift it 50 m at constant speed, while overcoming a frictional force of 115 N, is 710.37 kJ.
To calculate the work done on a 1,438-kg elevator car by its cable to lift it 50 m at constant speed, while overcoming friction, we use the following information:
- Elevator mass (m) = 1,438 kg
- Distance lifted (d) = 50 m
- Friction force (f) = 115 N
- Acceleration due to gravity (g) = 9.8 m/s2
The force exerted by the cable must counteract both the gravitational force and the frictional force. The gravitational force is
Fg = m x g,
and the total force (Ftotal) the cable must exert is
Fg + f.
Calculate Gravitational Force:
Fg = mg
= 1,438 kg x 9.8 m/s2
= 14,092.4 N
Total Force:
Ftotal = Fg + f
= 14,092.4 N + 115 N
= 14,207.4 N
Work Done by the Cable:
W = Ftotal x d
= 14,207.4 N x 50 m
= 710,370 J
Now, converting joules to kilojoules:
W = 710,370 J
(1 kJ / 1,000 J)
= 710.37 kJ
Therefore, the work done by the cable to lift the elevator car 50 m, overcoming friction, is 710.37 kJ.
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