Answer:
A) 31 kJ
B) Â 1.92 KJ
C) 40 , Â 2.48
Explanation:
weight of person ( m ) = 79 kg
height of jump ( h ) = 0.510 m
Compression of joint material ( d ) = 1.30 cm ≈  0.013 m
A) calculate the force
Fd = mgh
F = mgh / d
W = mg
F(net) = W + F Â = mg ( 1 + [tex]\frac{h}{d} )[/tex]
     =  79 * 9.81 ( 1 + (0.51 / 0.013) )
     = 774.99 ( 40.231 ) ≈ 31 KJ
B) calculate the force when the stopping distance = 0.345 m
d = 0.345 m
Fd = mgh  hence  F = mgh / d
F(net) = W + F Â = mg ( 1 + [tex]\frac{h}{d} )[/tex]
     =  79 * 9.81 ( 1 + (0.51 / 0.345) )
     = 774.99 ( 2.478 ) = 1.92 KJ
C) Ratio of force in part a with weight of person
= Â 31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40
 Ratio of force in part b with weight of person
= 1920 / 774.99 = 2.48