The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm
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In order to accelerate the dragster at a speed 
, its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
and the sign is negative because the frictional force acts against the direction of motion of the dragster.
This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is
:
So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
And since 1 horsepower is equal to 746 W, we can rewrite the power as
however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
and the sign is negative because the frictional force acts against the direction of motion of the dragster.
This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is
So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
And since 1 horsepower is equal to 746 W, we can rewrite the power as