An electric lift with a maximum load of $2000 ext{ kg}$ (lift+passengers) is moving up with a constant speed

Question

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

Accepted Answer

1. **Identify the Forces:** The motor must exert an upward force ($F$) to overcome the gravitational force (weight, $mg$) and the opposing frictional force ($f$). 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 ($F$):**
   $$ F = mg + f $$
   $$ F = (2000	ext{ kg} 	imes 10	ext{ m s}^{-2}) + 3000	ext{ N} $$
   $$ F = 20000	ext{ N} + 3000	ext{ N} = 23000	ext{ N} $$
3. **Calculate Power ($P$):** Power is the product of the force and the velocity ($P = \mathbf{F} \cdot \mathbf{v}$).
   $$ P = F 	imes v $$
   $$ P = 23000	ext{ N} 	imes 1.5	ext{ m s}^{-1} $$
   $$ P = 34500	ext{ W} $$
   [Class 11 Physics, Ch 5, Example 5.10]

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