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NEET PHYSICSEasy

An electric lift with a maximum load of 2000 kg2000\text{ kg} (lift+passengers) is moving up with a constant speed of 1.5 m s−11.5\text{ m s}^{-1}. The frictional force opposing the motion is 3000 N3000\text{ N}. The minimum power delivered by the motor to the lift in watts is: (Take g=10 m s−2g=10\text{ m s}^{-2})

A

23500

B

23000

C

20000

D

34500

Step-by-Step Solution

  1. Identify the Forces: The motor must exert an upward force (FF) to overcome the gravitational force (weight, mgmg) and the opposing frictional force (ff). Since the lift moves at a constant speed, the net force is zero, meaning the upward force balances the downward forces.
  2. Calculate Total Upward Force (FF): F=mg+fF = mg + f F=(2000 kg×10 m s−2)+3000 NF = (2000\text{ kg} \times 10\text{ m s}^{-2}) + 3000\text{ N} F=20000 N+3000 N=23000 NF = 20000\text{ N} + 3000\text{ N} = 23000\text{ N}
  3. Calculate Power (PP): Power is the product of the force and the velocity (P=F⋅vP = \mathbf{F} \cdot \mathbf{v}). P=F×vP = F \times v P=23000 N×1.5 m s−1P = 23000\text{ N} \times 1.5\text{ m s}^{-1} P=34500 WP = 34500\text{ W} [Class 11 Physics, Ch 5, Example 5.10]
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