Question

Refer to Interactive Solution 6.59 for a review of the approach taken in problems such as this one. A 53.0-kg person jumps from rest off a 4.40-m-high tower straight down into the water. Neglect air resistance during the descent. She comes to rest 1.40 m under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is non-conservative.

Answer 1

  MGH = MGH + Wnc
(53.0)(9.8)(4.40) = (53.0)(9.8)(-1.40) + Wnc
2285 = -727 + wnc
3012 = Wnc

W=FD
3012= F(1.4)
3012/1.4 = F
2151 = F

Answer 2

Consider B to C motion inside water ::

Vb = 9.29 m/s

Vc = 0 m/s

d = 1.40 m

Using Kinematics equation ::

Vc^2 + Vb^2 + 2 a d

0^2 = 9.29^2 + 2 a (1.40)

0 = 86.3 + 2.8 a

a = -30.82 m/s²

average force = mass x acceleration

F = (53 kg) (30.82) = 1633.5 N

F = 1633.5 N