An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 
. The frictional force opposing the motion is 4000 N. What is minimum power delivered by the motor to the elevator?
Text Solution
Verified by ExpertsThe correct answer is:
B
: Here, m = 1800 kg
Frictional force, f = 4000 N
Uniform speed, v = 2m s -1
Downward force on elevator is
F = mg+f
= 
The motor must supply enough power to balance this force. Hence,
= 44000 W = 44 ×10 3 W = 44 kW
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
A body of mass m accelerates uniformly from rest to in time . As a function of time t, the instan…
A man is riding on a cycle with velocity 7.2 km/hr up a hill having a slope 1 in 20. The total mass…
A 12 HP motor has to be operated 8 hours/day. How much will it cost at the rate of 50 paisa/kWh in …
A motor boat is travelling with a speed of 3.0 m/sec. If the force on it due to water flow is 500 N…
An electric motor exerts a force of 40 N on a cable and pulls it by a distance of 30 m in one minut…
An electric motor creates a tension of 4500 newton in a hoisting cable and reels it in at the rate …