Understanding the Electric Lift Problem

This problem asks us to find the instantaneous power delivered by the motor of an electric lift at a specific moment when it's moving upwards at a certain speed. We are given the lift's maximum load (mass), its upward acceleration, the opposing frictional force, and the speed at which we need to calculate the power. We also know the acceleration due to gravity, g.

Key Information Provided:

• Mass of the lift plus passengers, m = $2000\text{ kg}$
• Upward acceleration, a = $0.2\text{ ms}^{-2}$
• Frictional force opposing motion, f = $3000\text{ N}$
• Velocity at the instant of interest, v = $1.5\text{ ms}^{-1}$
• Acceleration due to gravity, g = $10\text{ ms}^{-2}$

Calculating Forces Acting on the Lift

To find the power, we first need to determine the force the motor must exert. Several forces act on the lift:

1. Weight (W): This force acts downwards due to gravity. It's calculated as mass times gravity. $W = m \times g$ $W = 2000\text{ kg} \times 10\text{ ms}^{-2} = 20000\text{ N}$
2. Frictional Force (f): This force opposes the motion, so it also acts downwards since the lift is moving upwards. $f = 3000\text{ N}$
3. Tension (T): This is the upward force exerted by the motor via the cable. This is the force we need to find to calculate power.

Applying Newton's Second Law

Newton's Second Law states that the net force acting on an object is equal to its mass times its acceleration (ΣF = ma). For the lift moving upwards:

The upward forces are balanced by the downward forces plus the force required for acceleration.

Net upward force = Tension - Weight - Frictional Force

$T - W - f = ma$

Now, we plug in the known values:

$T - 20000\text{ N} - 3000\text{ N} = (2000\text{ kg}) \times (0.2\text{ ms}^{-2})$ $T - 23000\text{ N} = 400\text{ N}$

To find the tension (the force the motor provides), we rearrange the equation:

$T = 400\text{ N} + 23000\text{ N}$ $T = 23400\text{ N}$

So, the motor needs to exert an upward force of $23400\text{ N}$ at this instant.

Calculating Instantaneous Power

Power is the rate at which work is done. For an object moving with velocity v under a force F, the instantaneous power P is given by the product of the force and the velocity in the direction of the force.

In this case, the force delivered by the motor is the tension T, and the velocity is v.

$P = T \times v$

Substituting the values we found:

$P = 23400\text{ N} \times 1.5\text{ ms}^{-1}$ $P = 35100\text{ W}$

The unit of power is Watts (W).

Conclusion

The instantaneous power delivered by the motor when the lift reaches a speed of $1.5\text{ ms}^{-1}$ is $35100\text{ W}$.