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Work and Power question

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data

    A 1100kg car can accelerate from 0-60kph in 6.5 seconds. With constant engine power, what is the steepest hill the car could climb at 20 kph.


    2. Relevant equations

    P=W/T
    Sliding force of the slope = 1100g(sinTheta)
    A= 16.67/6.5 --> 2.56m/s^2
    SF=MA

    3. The attempt at a solution

    I found the sum of the forces (1100 * 2.56) = 2816
    and set the equation up as ... 2816=1100g(sinTheta)
    though I know this is wrong because the answer is 23. I dont know how to incorperate that fact that the car is climbing at 20kph.
     
  2. jcsd
  3. Mar 24, 2012 #2

    gneill

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    Staff: Mentor

    The first thing to establish is the power output of the engine given its performance of 0-60kph in 6.5s. (hint: what's the energy expenditure over time?)

    The next bit will be a bit more tricky, establishing the power required to climb a slope at a given speed. Where will energy "go" when you climb a slope?
     
  4. Mar 24, 2012 #3
    How do I find work to find power if there are no distances given in the question ? I know I can find the sum of the forces through MA, but do I need that?
     
  5. Mar 24, 2012 #4

    gneill

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    Power is energy delivered per unit time. How much energy did the car gain over the given time when it went from 0 to 60 kph in 6.5 seconds?

    You don't necessarily require the forces involved if you consider the energy expenditure versus time to climb the slope. Where is the energy going? If the speed along the incline is known, how much energy is being expended per unit time?
     
  6. Mar 24, 2012 #5
    It is becoming heat no? But again how would I find the energy the car has gained or has ?
     
  7. Mar 24, 2012 #6

    gneill

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    You can ignore friction (so no heat). Where does kinetic energy "go" when something gains height in a gravitational field?
     
  8. Mar 24, 2012 #7
    It gains potential energy.
     
  9. Mar 24, 2012 #8

    gneill

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    Yes. So relate the speed of the car up the incline to the rate of gain of potential energy...
     
  10. Mar 24, 2012 #9
    I have no Idea how to solve this... 2816=1100g(sinTheta) is that close to how you solve it ? I know im just missing the climbing velocity component i don't know how to incorporate.
     
  11. Mar 24, 2012 #10

    gneill

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    Before you tackle the slope, have you determined the power output of the car engine? You will need to solve this portion of the problem before you can tackle the incline.
     
  12. Mar 24, 2012 #11
    the car is accelerating at 2.56m/s, and the car weighs 1100kg, if i multiply those I get the sum of the forces on the engine 2820.5N. How may that help me find the power ? How do I find the power ?
     
  13. Mar 24, 2012 #12

    gneill

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    Rather than thinking in terms of kinematic details like acceleration and force, think in terms of energy. How much energy did the car gain in going from 0 to 60 kph? In what time did it acquire this energy? So what was the power (energy/time)?
     
  14. Mar 25, 2012 #13
    so would i do 1/2mv^2... (0.5)*(1100)*(16.67)^2/6.5 ? then multiply that by 5.55 (20kph) to get the power of the car going up the slope ?
     
  15. Mar 25, 2012 #14

    gneill

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    Yes, that will give you the power that the car produces. The kinetic energy that the car achieves is equal to the work done, and it was done over time 6.5s.
    No, you got the power when you divided the change in KE by time.

    As I hinted earlier, dealing with the slope is a bit trickier. There are a couple of possible approaches: (1) an "energy approach" where we use energy conservation to see at what rate the car can gain height using its power, the energy going into gravitational potential energy; (2) a force balance approach, where the force required to maintain the car's speed up the slope is considered.

    I you go with the second approach, then what can you say about the net force acting on the car parallel to the slope as it climbs at a constant speed? Also, what is the relationship between power, force, and speed?
     
  16. Mar 25, 2012 #15
    so.. The power of the engine = 23'514watts, and this is constant as it climbs the hill therefor the opposing force (sliding) is = to it.

    though when I set up 23514= 1100g*(Sinθ)... i still dont get the answer, I get an error.
     
  17. Mar 25, 2012 #16
    And for the relationship as force increases, so does power, and the more power there is, the greater the speed the car has.
     
  18. Mar 25, 2012 #17

    gneill

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    That's because your 23514 number has units of power, not force. You have yet to determine what force that power delivers at 20 kph.
     
  19. Mar 25, 2012 #18

    gneill

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    There is a specific mathematical relationship that relates the power delivered by a force to an object moving at a given velocity. Hint: The unit for power is the Watt (W) which is also J/s, which is also N*m/s :smile:
     
  20. Mar 25, 2012 #19
    So I would need to divide by m and multiply by s ? How would I convert the power to N. And how do i convert 20kph to a force ?
     
  21. Mar 25, 2012 #20
    I did it
     
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