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Efficiencey of a jack

  1. May 5, 2009 #1
    The end of the lever of a tire jack travels 7 m for every centimeter that it lifts the car. If the car has a mass of 1000 kg and a force of 50 N is needed to lift the car, what is the efficiency of the jack


    Efficiency = useful work output/total work input
    w = F x d, where w is work, F is force, and d is distance. (If given mass instead of force, force is found by multiplying mass (in kg) by 9.8 m/sec2 (acceleration of gravity on earth)



    the useful work input would be .05 correct? because you would take the 50 N the force times the distance .001 m but i don't know how to find the work output as it doesn't tell a force.. am i missing something?
     
  2. jcsd
  3. May 5, 2009 #2

    mgb_phys

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    You are almost there !
    Efficiency = work out / work in
    work = force * distance

    So Efficiency = (weight_car * distance car) / (force_jack * distance jack)
     
  4. May 5, 2009 #3
    but i don't know what the force on the jack is...
     
  5. May 5, 2009 #4

    mgb_phys

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    It says 50N
     
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