If the downward velocity of the leg is 1.0m/s when the jogger's foot hits the ground, and if the leg stops...
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If the downward velocity of the leg is 1.0m/s when the jogger's foot hits the ground, and if the leg stops...

[From: ] [author: ] [Date: 12-08-13] [Hit: ]
0^2 +2a(.a = -1/((2)(.a = -1/(.A 75kg jogger as a force of 75 x g =75 x 9,......
Severe stresses can be produced in the joints by jogging on hard surfaces or with insufficiently padded shoes. If the downward velocity of the leg is 1.0 m/sec when the jogger's foot hits the ground, and if the leg stops in a distance of 2.0 cm, calculate the force in the ankle joint. The mass of the jogger's leg is 12 kg. (You may neglect the weight of the rest of the jogger's body.) Compare this force with the weight of a 75-kg jogger.

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F=ma

calculate acceleration.

Vf^2 = Vi^2 + 2ad

where Vf = final velocity =0
Vi - initial velocity = 1.0 m/s
d = distance = 2cm = .02m

0 = 1.0^2 +2a(.02)

a = -1/((2)(.02)

a = -1/(.04)
a = -25m/s^2

F=ma
F = 12kg(-25m/s^2)

A 75kg jogger as a force of 75 x g =75 x 9,8
or 735 N
F = 300 N
1
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