Question

Consider a skydiver of mass 75 kg who jumps from an airplane flying at an altitude of 1,900 m. With her parachute open, her terminal velocity is 8.1 m/s.

(a) What is the work done by gravity?
J

(b) What is the average force of air drag during the course of her jump? (Enter the magnitude of the force.)
N

Answer 1

a) Workdone by gravity, W_gravity = m*g*h*cos(0)

= m*g*h

= 75*9.8*1900

= 1396500 J <<<<<<<<<<------------Answer

b) use net Workdone = change in kinetic energy

W_gravity + W_air = KEf - KEi

1396500 + W_air = (1/2)*m*vt^2 - 0

W_air = (1/2)*m*vt^2 - 1396500

W_air = (1/2)*75*8.1^2 - 1396500

= -1394040 J

now use, W_air = F_air*h*cos(180)

W_air = -F_air*h

==> F_air = 1394040/1900

= 734 N <<<<<<<<<----------------Answer