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.
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
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
Refer to Interactive Solution 6.59 for a review of the approach taken in problems such as...
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