Asked by Paul
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 7.3m/s when the joggers foot hits the ground. If the leg stops in a distance of 1.1 cm. Calculate the force on the ankle joint. The mass of the joggers leg is 9.5 kg
Answers
Answered by
Damon
stops in .011 meter
average speed during deacceleration = 7.3/2 = 3.65 m/s
so time to stop = .011/3.65 = .00301 second
momentum change = 9.5 * 7.3 = 69.4 kg m/s
so F = d (mv)/dt = 69.4 kg m/s / .00301 s
= 23012 kg m/s^2 or Newtons
Note - I used Newtons original second law which is Force = rate of change of momentum. For constant mass this is F = m a
average speed during deacceleration = 7.3/2 = 3.65 m/s
so time to stop = .011/3.65 = .00301 second
momentum change = 9.5 * 7.3 = 69.4 kg m/s
so F = d (mv)/dt = 69.4 kg m/s / .00301 s
= 23012 kg m/s^2 or Newtons
Note - I used Newtons original second law which is Force = rate of change of momentum. For constant mass this is F = m a
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