How Efficient Is a Tire Jack Lifting a Car with a 50N Force?

[tuffshorty]

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?

 

[mgb_phys]

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

So Efficiency = (weight_car * distance car) / (force_jack * distance jack)

 

[tuffshorty]

but i don't know what the force on the jack is...

 

[mgb_phys]

It says 50N