Minimum power of an electric lift – Rankers Physics

Power: Practice Problem & Solution

An electric lift with a maximum load of $2000\text{ kg}$ (lift + passengers) is moving up with a constant speed of $1.5\text{ ms}^{-1}$. The frictional force opposing the motion is $3000\text{ N}$. The minimum power delivered by the motor to the lift in watts is : ($g = 10\text{ ms}^{-2}$) (2022)
$23500$
$23000$
$20000$
$34500$

Solution Explained:

To solve this problem, we apply the core principles of Power. Understanding the underlying formula is key to arriving at the correct answer below:

Total upward force required equals the sum of weight and frictional force: $F = mg + f = (2000 \times 10) + 3000 = 23000\text{ N}$. Minimum power $P = F \cdot v = 23000 \times 1.5 = 34500\text{ W}$.

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