Question
(a) Find the useful power output of an elevator motor that lifts a 2500-kg load a height of 35.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg—so that only 2500 kg is raised in height, but the full 10,000 kg is accelerated. (b) What does it cost, if electricity is $0.0900 per  kW⋅ hr\textrm{ kW} \cdot \textrm{ hr}?
Question by OpenStax is licensed under CC BY 4.0
Final Answer

a) 1.36×105 W1.36 \times 10^5 \textrm{ W}

b) 4.08 cents

Note: For part (b) the video calculates the cost per hour, rather than the cost to raise the elevator for 12 seconds. Multiplying the cost per hour by 12 seconds, and by 1 hr per 3600 seconds, gives the correct final answer of 4.08 cents. The calculator screenshot shows this final calculation.

Solution video

OpenStax College Physics, Chapter 7, Problem 39 (Problems & Exercises)

OpenStax College Physics, Chapter 7, Problem 39 (PE) video thumbnail

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Calculator Screenshots

  • OpenStax College Physics, Chapter 7, Problem 39 (PE) calculator screenshot 1
  • OpenStax College Physics, Chapter 7, Problem 39 (PE) calculator screenshot 2
Video Transcript
This is College Physics Answers with Shaun Dychko. The system here consists of an elevator that has a mass of 2500 kilograms and a counterweight on the other end of the pulley with mass of 7500 kilograms because we're told that the total weight of the system is -- total mass I should say, is 10,000 kilograms. So this is 10,000 minus the 2,500 of the elevator. So the elevator goes upwards a height of 35 meters and in order for that to happen this counterweight must be going downwards the same amount. So the change in height of the counterweight is negative 35 meters and the change in height of the elevator is positive 35 meters where we're taking upwards to be the positive direction. So to find the power output of the motor we need to figure out the energy consumed by the motor first of all. Then we'll divide that by the 12 seconds time it takes to raise the elevator 12 or 35 meters, and also to impart some speed to it because the elevator and counterweight will have a final speed of 4 meters per second. So, the initial kinetic and potential energy plus the energy input from the motor has to equal the total final energy, kinetic and potential. So, the initial kinetic and potential energy are both zero because we take the initial position to be the reference level so it has no height there. The initial kinetic energy is zero because it is not moving. Then on the right hand side we substitute one half total mass times final speed squared for the kinetic energy, and then for the potential energy we have two potential energies. We have the change of potential energy of the elevator which is the mass of the elevator times g times it's change in height. Then we have the potential energy of the counterweight. So, m w g times the change in height of the weight. We can factor out the g from these two terms just to make this look a little bit neater. Then divide that by t to get the power output of the motor. We're taking the absolute value of the work here of the motor because we don't really care what the sign is. We just want the magnitude to find the power. So, we have one half times 10,000 kilograms total mass, times 4 meters per second squared, plus 9.8 meters per second squared times 2,500 kilograms mass of the elevator, multiplied by positive 35 meters. Then add to that 7,500 kilograms mass of the counterweight times negative 35 meters and that's a negative because the counterweight is going down and divide all that by 12 seconds. We get 1.36 times ten to the six watts is the power output of the motor. The cost of this will be that multiplied by one kilowatt per 1,000 watts, so we're dividing by 1,000 in other words, and then we multiply by 0.09 dollars per kilowatt hour. So the kilowatts cancel and we're left with dollars per hour. That's 12.26 dollars per hour is the cost if it was operating and continuously lifting for an entire hour. This is what it would cost.

Comments

Part (b) asks for the cost to raise the elevator the height in part (a) - which means that you need to multiply the cost per unit time by the time (12 s) - the final answer is 4.087 cents.

I understand the kinetic and potential energies of the elevator, but I'm a bit confused why the counterbalance potential is included. If the counter balance potential energy, wouldn't we have to include it's final kinetic energy too?

Perhaps it's because it's talking about the system as a whole. The kinetic energy of the entire system, which does include both elevator and counterweight. However, if the final Counterweight KE is 0, I could see why it was left out. Over all - That's how my mind has diagnosed it. I could be wrong. --Shaun?

Thank you for the question Russell. The kinetic energy of the counterweight is definitely included. At 1:41 I mention that the mass in the kinetic energy term is the total mass of the system, which is to say the mass of the counterweight plus the mass of the elevator with occupants. The potential energy has two terms - one for each mass - since the masses move differently. The counterweight moves negative 35.0 m, whereas the elevator moves positive 35.0 m. Considering kinetic energy, however, they each move with the same speed so their respective kinetic energy terms can be collected into one term. In other words: 12mEvf2+12mwvf2=12(mE+mw)vf2=12mTvf2\dfrac{1}{2}m_\textrm{E}v_f^2 + \dfrac{1}{2}m_\textrm{w}v_f^2 = \dfrac{1}{2}(m_\textrm{E}+m_\textrm{w})v_f^2 = \dfrac{1}{2}m_\textrm{T}v_f^2, where the final expression has mTm_\textrm{T} with the subscript T for total mass which is what's shown in the video.
Hope this helps,
Shaun