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What are the steps to solve a work problem?

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  1. Oct 20, 2014 #1
    I know there isn't a "set in stone" step by step procedure to solve work problems, but if someone could give me a basic procedure it would be greatly appreaciated. When ever I read a work problem, unless it is a simple "plug in the value" works equation problem, I do not understand how to start off the problem. Maybe its just that I don't understand the concept?
    For example, can you make a procedure that fits this problem:
    "Calculate the average power output necessary for a 55.8 kg person to run up a 12.0 m long hillside, which is inclined at 25.0° above the horizontal, in 3.00 s. Express your answer in horsepower."
     
  2. jcsd
  3. Oct 21, 2014 #2

    D H

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    • Determine what the question is asking for. In this case it's obvious: The question is asking for "average power".
    • Know your units. In this case, power is energy per unit time.
    • Filter out the irrelevant information. In this case, the person is running up a hill. From a biomechanical perspective, it takes a good deal of energy to run on flat ground for 26 miles. From a physics perspective, how humans run is irrelevant.
    • Determine what is relevant. In this case, it's vertical distance.
    • There are some things you just have to know. In this case, it's the relation between change in altitude and energy.
    • Watch out for tricks. In this case, you're not quite finished when you get an answer in joules. You need to convert to horsepower.
    • Answer the question!
     
  4. Oct 21, 2014 #3
    This is one area of physics that I think gets easier when you get to more advanced levels, where such a procedure exists.

    But for basic physics I would say to try this. Just quickly ask yourself what is doing the work, and what is it doing the work on. Then just quickly think about W=F*d to predict the sign of the work.
    Then to actually find the value you can either use W=F*d or W=-ΔU where U is potential energy. (in case you don't know ΔU = Ufinal - Uinitial)

    I saw the other thread you made in which you were asking about when to use PE or KE (which I'll label U and T). Notice I didn't write an equation involving T (kinetic energy) above, because the cases in which you substitute T for work come from W=-ΔU too. Usually this is because the problem will be a situation in which ΔU=-ΔT.
    So then you can see that if W=-ΔU then for that specific problem work must also be W=ΔT and you can pick whichever one is easier to solve.

    However, in this problem you do not have a free-body moving under the influence of only external forces (because the human is generating internal forces to run) so ΔU is not equal to -ΔT and therefore you cannot use kinetic energy to solve this problem, and you use potential energy U instead.

    Also, if you are getting hung up on signs, just be aware that a lot of teachers and even textbooks get lazy with their signs for work and sometimes get the signs wrong. Just a warning so you don't waste too much time troubling over work signs like I remember doing.
     
  5. Oct 21, 2014 #4

    Simon Bridge

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    Related to:
    https://www.physicsforums.com/threa...in-kinetic-energy-or-potential-energy.777302/
    ... it is important to understand what you are being asked for, yes.
    Perhaps you should work through some examples, showing us how you are thinking about each problem?

    ... actually, if a change in speed were given, then the runner would have to generate the power, internally, for that too; and you may have to use KE for the calculation as well. They key is that the total change in energy is zero.

    Sometimes the question has to be read carefully:
    In the example, the question does not care about the work to change speed, or pant or keep running or anything else, - only the work to get to the top of the hill.
     
    Last edited: Oct 21, 2014
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