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Efficiencey of a jack

  • Thread starter tuffshorty
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
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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)
 
  • #3
11
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but i don't know what the force on the jack is...
 
  • #4
mgb_phys
Science Advisor
Homework Helper
7,774
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It says 50N
 

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