"rocket man" has a propulsion unit strapped to his back. he starts from rest on the ground, fires the unit, and accelerates straight upward. at a height of 15 m, his speed is 4.8 m/s. his mass, including the propulsion unit, has the approximately constant value of 135 kg. find the work done by the force generated by the propulsion unit. (neglect air resistance.)

For the work-energy theorem, the work done by the propulsion unit should be equal to the total mechanical energy E gained by the rocket man:

where K is the kinetic energy gained by the rocket man, while U is its gravitational potential energy at height h=15 m, where his speed is v=4.8 m/s. Expliciting the terms, we find the work:

What formula  this one  (x_/

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j