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Homework Help: Mechanics - Work-Energy Principle Question

  1. Mar 14, 2014 #1
    1. The problem statement, all variables and given/known data

    a) A car of mass 1050kg moves along a straight horizontal road with its engine working at a constant rate of 25kW. Its speed at a point A on the road is 12ms-1. Assuming that there is no resistance to motion, calculate the time taken for the car to travel from A until it reaches a speed of 20ms-1.

    b) Assume now that there is a constant resistance to motion and that the car's engine continues to work at 25kW. It takes 10.7s for the car's speed to increase from 12ms-1 to 20ms-1. During this time the car travels 179m. Calculate the work done against resistance and hence find the magnitude of the resistance.

    c) Later the car moves up a straight hill, inclined at 2° to the horizontal. The engine works at 25kW as before, and there is a constant resistance of the same magnitude as before. The car travels a distance of 393m while its speed increases from 12ms-1 to 20ms-1. Calculate the time taken by the car to travel this distance.

    2. Relevant equations

    I'm lost at part b). I don't know what i am missing with my understanding.

    3. The attempt at a solution

    Given that gravitational force is conservative, the work-energy principle can be used.
    [tex]P=\frac{E}{t}=\frac{m\left ( v^{2}-u^{2} \right )}{2t}[/tex]
    [tex]t=\frac{m\left ( v^{2}-u^{2} \right )}{2P}[/tex]
    [tex]t=\frac{1050\left ( 20^{2}-12^{2} \right )}{2\times 25000}[/tex]
    This answer is confirmed correct

    [itex]P=Fv[/itex] so [itex]F=\frac{P}{v}[/itex]. [itex]F=ma[/itex] so with a resistive force
    Given that work done [itex]W=Fd[/itex] and the net force in the system is now [itex]\frac{P}{v}-R[/itex] then
    [tex]W=\left ( \frac{P}{v}-R \right )d[/tex]
    So we need the magnitude of [itex]R[/itex] before we can find work done. To find [itex]R[/itex] we need to know the acceleration.
    Now can find [itex]R[/itex]
    [tex]R=\frac{25000}{12}-1050\left ( 0.748 \right )[/tex]
    This is incorrect. Don't know what i'm missing here. Any help would be great. Cheers
  2. jcsd
  3. Mar 14, 2014 #2


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    There is no need to assume that acceleration is constant .
  4. Mar 14, 2014 #3
    How else can i find [itex]R[/itex] without knowing the acceleration? And wouldn't the acceleration be constant given that the resistive force is constant?
  5. Mar 14, 2014 #4


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    It may be constant, but you're making a assumption about that.

    The resistive force is constant, but is the NET force constant?

    How much work is done by the resistive force?

    What is the net work done on the car?
  6. Mar 15, 2014 #5
    If the engine continues to work at a constant rate then won't the force generated by the engine also be constant? Therefore the net force is constant.

    I don't see how i can find the work done by the resistive force because i can't use the work-energy principle as frictional forces are not conservative.
  7. Mar 15, 2014 #6


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    You can determine the Net work done on he system.

    Then, you can set up an equation with the resistive force being the only unknown.
  8. Mar 18, 2014 #7
    How do i find the net work done if i don't know the net force? I can't used work-energy principle as friction is not conservative.
    If i just use [itex]F=\frac{P}{v}[/itex] then aren't i assuming [itex]v[/itex] is constant? Clearly it's not as it's increasing over the distance.
    Definitely not getting this :(
  9. Mar 18, 2014 #8
    Hey Ant, I think you are going about this with a bit of calculus pretty easily. Considering you have a constant resisting force and your engine is providing a constant power, I think it is safe to say that you WILL have a constant acceleration. So, remember that the derivative of your work and energy equation with respect to time will give you your power equation.

    So, if you set it up,

    1/2mv^2 + ∑W = 1/2mv^2

    So, the two forces doing work on your object are the force coming from the engine which is working WITH the direction of motion, and the resisting force which is working against it (that means watch your signs). Both of those are unknowns, and after you set that up, take the time derivative of the entire equation and you will be able to work with values that you already know. Remember the chain rule when you take the derivative of the kinetic energy, so that your acceleration pops up.

    After that, I would consider working backwards from that point keeping in mind that:

    P = W/t (when the power is constant, which is looks like it is)


    W = F*d

    I hope that helped!
  10. Mar 18, 2014 #9


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    The Work-Energy theorem does not require the forces to be conservative.

    The engine power is constant, so it's easy to find the work it does in 10.7 seconds.

    You can calculate the change in Kinetic Energy.

    The difference must be equal to the work done by the resistive force, R.

    You also know the distance traveled during the 10.7 seconds.
  11. Mar 19, 2014 #10
    I thought W-E theorum does require conservative forces. Better check that out again.

    For the work done by the engine:
    [tex]W_{E}=Fd=\frac{P}{v}d=\frac{25000}{12}\cdot 179=372916J[/tex]
    So the net work done:
    [tex]W_{N}=\frac{1}{2}m\left ( v^{2}-u^{2} \right )=\frac{1050}{2}\left ( 20^{2}-12^{2} \right )=134400J[/tex]
    Work done by resistance:
    Magnitude of resistive force:
    This isn't right, comparing with the answers.
  12. Mar 19, 2014 #11
    That first formula is incorrect. The force in not constant so you cannot calculate the work done as

    But you don't need to go through this. You know the power and the time. What is the definition of power?

    And you don't need conservative forces to apply work-energy theorem. It works for friction forces too.
    Or engine forces, as here.
    The conservation of energy requires conservative forces.
  13. Mar 19, 2014 #12
    Ok so for the energy burn from the engine we have
    [tex]W_{E}=Pt=25000\times 10.7=267500J[/tex]
    And for the net energy burn of the system we have
    [tex]W_{N}=\frac{1}{2}m\left ( v^{2}-u^{2} \right )=\frac{1050}{2}\left ( 20^{2}-12^{2} \right )=134400J[/tex]
    The difference is the work done by the resistive force
    [tex]F_{R}d=133100\; \Rightarrow \; F_{R}=\frac{133100}{179}=744N[/tex]
  14. Mar 19, 2014 #13
    This looks OK.
  15. Apr 17, 2014 #14


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    Work Energy Principle

    Hi Ant.
    This problem works out fine. Many people’s problem is that they think analysing acceleration because they assume F=ma is relevant. It is not relevant. WE principle is all about energies at certain times or positions (distances/heights). And that is it.
    Don’t get involved in what is happening between these energy ‘points’. No assumptions should be made that the access are linear or anything else. This is all about velocity changes, distance changes (against resistive forces) and/or height changes. Energy sources balanced against energy sinks if you will. In this case gravity is not the energy source. The engine is. And it provides energy at a rate of 25000J/s. power x time = energy supplied.
    Energy sinks are increase in KE, increase in GPE and work done against resistive forces over particular distances. That’s all. Forget conservative arguments, red herring.

    Happy to discuss
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