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

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The discussion revolves around calculating the efficiency of a tire jack lifting a car using a 50 N force. The efficiency formula is defined as useful work output divided by total work input. The work input is calculated using the force of 50 N multiplied by the distance lifted, which is 0.001 m. Participants are attempting to determine the work output, which involves the weight of the car and the distance it is lifted, but there is confusion regarding the force applied by the jack. Clarification on these calculations is necessary to finalize the efficiency assessment.
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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?
 
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Efficiency = work out / work in
work = force * distance

So Efficiency = (weight_car * distance car) / (force_jack * distance jack)
 
but i don't know what the force on the jack is...
 
It says 50N
 

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