1. A 77 kg athlete is sprinting at 9.3 m/s. During ground contact of one stride
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Question
1. A 77 kg athlete is sprinting at 9.3 m/s. During ground contact of one stride the athlete's Achilles tendon stretches 9 mm. How much strain energy is stored in the tendon if it has a spring constant of 320 kN/m? A 62 kg pole vaulter falls from a peak height of 6.25 m and lands on a thick mat. At first contact with the mat, his center of gravity is only 1.30 m high. During impact the mat compresses to 0.65 m high What average force was exerted by the mat on the pole vaulter during impact? NOTE: Ignore the change in strain energy (ASE-0) during the impact. 2. A 63 kg woman is hopping up and down using only her ankles. hop she generates an average power of 942.33 W over 180 ms while raising her center of mass 12 centimeters. Answer the following questions using the principles of work, energy & power. 3. During the propulsive phase of the What was the average work done on the woman by the ground reaction force during the propulsive phase? What was the average ground reaction force experience by the woman during the propulsive phase? What was the woman's kinetic energy at takeoff? How high was the woman's hop? Use the Law of Conservation of Energy a. b. c. d.Explanation / Answer
1) the tendon here behaves like a spring
Energy stored in spring =1/2Kx²
K=320KN/m =320*10³ N/m
X=9mm=9*10—³ m
Energy stored=0.5*320*10³*(9*10—³)²
Energy=0.5*320*81*10^-6*10³
Energy=12960*10—³
Energy =12.960
2);when she was at the highet of 1.30 m after falling, velocity, from equation of motion is
V²=U² +2gh
As u=0
V²=2*9.8*6.25=122.5
V=11.06 m/s...........(1)
So momentum =mv=11.06*62=686.21
When she is at the highet of 0.65 velocity =0 m/s and momentum =0.
From the time of contact to the time of stoppage we have
V²=u² +2as
As v=0
U²=2as
Put U=v from 1 we get
11.06² =2*a*(1.30-0.65)
122.08=2a*0.65
a=122.08/1.30 =93.9 m/s²
From another equation of motion we have
V'=u' - at
As v'=final velocity =0 and u'=initial velocity =v from 1
We get
11.06=93.9*t
t=0.117 seconds. This is the total time of contact
So from impulse momentum equation we have
Average force=change in momentum /time
Average force=686.21 - 0/0.117
Average force =5865 N
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