Answer:
a) W = ΔK.E + ΔP.E = (mv²/2) - mgd
b) Power supplied by the drag force after the skydiver has reached terminal velocity = mgv
Explanation:
Using the work energy theorem, the work done by the drag force while the skydiver moves through a distance d is equal to the change in kinetic energy and potential energy of the body.
The potential energy of the skydiver drops by a magnitude of (mgd) falling through a distance of d.
ΔP.E = - mgd (skydiver loses this amount of potential energy)
But the skydiver falls from rest to gain a velocity of v after falling a distance of d vertically.
ΔK.E = (final kinetic energy at distance d) - (initial kinetic energy at rest)
Final kinetic energy at d = (1/2)(m)(v²) (the body is now falling with terminal velocity (v) at this point d)
Initial kinetic energy = 0 (since the skydiver was initially at rest)
ΔK.E = (1/2)(m)(v²) - 0 = (mv²/2)
W = ΔK.E + ΔP.E = (mv²/2) - mgd
b) At terminal velocity, the net force on the skydiver = 0
hence,
drag force = weight of the skydiver = mg
Power = F × v = mg × v = mgv
Hope this Helps!!!
