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AP Physics 1 - Elevator Motor Power Calculation

Created by @nathanedwards
 at November 1st 2023, 8:22:23 am.
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#power
#ap-physics-1
#elevator

AP Physics 1 Exam Question - Power

A 500 kg elevator is ascending at a constant speed of 2 m/s. The elevator motor exerts a force of 8000 N to lift the elevator.

  1. Calculate the power output of the elevator motor.

Solution:

Power (P) is defined as the rate at which work is done or energy is transferred. It can be calculated using the formula:

P=WΔt P = \frac{W}{\Delta t}

where P is power, W is work, and Δt is the time taken to do the work.

In this case, the power output of the elevator motor can be calculated by determining the work done by the motor in lifting the elevator over a specific time interval. Since the elevator is ascending at a constant speed, the work done is the product of the force exerted and the distance covered.

Force (F)=8000 N \text{Force (F)} = 8000 \ \text{N}
Speed (v)=2 m/s \text{Speed (v)} = 2 \ \text{m/s}
Δt=DistanceSpeed \text{Δt} = \frac{\text{Distance}}{\text{Speed}}

To find the distance covered, we can use the kinematic equation:

v2=u2+2ad v^2 = u^2 + 2a\text{d}

Since the elevator is ascending at a constant speed, the initial velocity (u) is equal to the final velocity (v), and the acceleration (a) is 0. Rearranging the equation, we have:

v2=2ad v^2 = 2a\text{d}
v22a=d \frac{\text{v}^2}{2a} = \text{d}

Substituting the given values, we have:

d=v22a=(2 m/s)22(0)=∞ \text{d} = \frac{\text{v}^2}{2a} = \frac{(2 \ \text{m/s})^2}{2(0)} = \infty

Since the distance covered is infinite (as the elevator continues to ascend), we can consider a specific time interval, such as 1 second.

Now, let's calculate the power output of the elevator motor:

Work (W)=Force×Distance=(8000 N)×(2 m/s)=16000 Joules \text{Work (W)} = \text{Force} \times \text{Distance} = (8000 \ \text{N}) \times (2 \ \text{m/s}) = 16000 \ \text{Joules}
Δt=1 second \text{Δt} = 1 \ \text{second}

Using the formula for Power:

P=WΔt=16000 J1 s=16000 Watts P = \frac{W}{\Delta t} = \frac{16000 \ \text{J}}{1 \ \text{s}} = 16000 \ \text{Watts}

Hence, the power output of the elevator motor is 16000 Watts.

Note: The given scenario assumes an idealized case where there is no friction or other resistive forces acting on the elevator. In practice, power losses due to friction and other factors would reduce the calculated power output.

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