An elevator can carry a maximum load of 1800 kg and move up with a constant speed. Class 9

An elevator can carry a maximum load of 1800 kg and move up with a constant speed. Class 9

Question 1.

An elevator can carry a maximum load of 1800 kg and move up with a constant speed of 2 m/s. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator. Class 9

Answer:

Load = mg = 1800 × 10 = 18000 N

Force of friction is f = 4000 N

Downward force on the elevator is F = mg + f = (1800 × 10) + 4000 = 22000 N

Question 2.

An engine pumps 400 kg of water through height of 10 m above the ground in 40 s. How much electric power is consumed by the engine? Class 9

Answer:

P = [latex]\frac{W}{t}[/latex] W = mgh = 400 × 10 × 10 = 40000 J

Power used by engine is P = 40000/40 = 1000 W