A3.5 kg block slides along a frictionless horizontal with a speed of 2.5 m/s. after sliding a distance of 5 m, the block makes a smooth transition to a frictionless ramp inclined at an angle of 50â—¦ to the horizontal. the acceleration of gravity is 9.81 m/s 2 .

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

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