Hello, I am preparing for my physics finals. I am working on the study guide and I am not sure if I solve the last problem correctly. Can someone please let me know if I did correctly solve it. If I did not solve correctly can you please guide me. Thank you in advance...
An athlete who weighs 800 N is able to raise his center of mass 0.500 m in a vertical jump.
(a) Compute the internal work done by the athlete’s leg muscles as he pushes off from the ground.
(b) Find the athlete’s speed as the feet leave the ground.
(c) Find the time during which the feet are in contact with the ground and the body accelerates upward, assuming that the center of mass moves through a distance of 0.400 m at constant acceleration.
(d) Calculate the average mechanical power produced.
solution:
a) work = force * distance
800 * 0.500= 400J
b) v= sqr (2*g*d)
sqr (2*9.8*0.500)
v= 3.13m/s^2
c) v = u + gt
3.13= 0 + 9.8t
t= 0.32 sec
d) Average Power (Watts) = √4.9 x body mass (kg) x √jump-reach score (m) x 9.8
Average Power = 2.2136 x 82 x 0.7071 x 9.8
Average Power = 1303.8 watts
An athlete who weighs 800 N is able to raise his center of mass 0.500 m in a vertical jump.
(a) Compute the internal work done by the athlete’s leg muscles as he pushes off from the ground.
(b) Find the athlete’s speed as the feet leave the ground.
(c) Find the time during which the feet are in contact with the ground and the body accelerates upward, assuming that the center of mass moves through a distance of 0.400 m at constant acceleration.
(d) Calculate the average mechanical power produced.
solution:
a) work = force * distance
800 * 0.500= 400J
b) v= sqr (2*g*d)
sqr (2*9.8*0.500)
v= 3.13m/s^2
c) v = u + gt
3.13= 0 + 9.8t
t= 0.32 sec
d) Average Power (Watts) = √4.9 x body mass (kg) x √jump-reach score (m) x 9.8
Average Power = 2.2136 x 82 x 0.7071 x 9.8
Average Power = 1303.8 watts
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a) 400J is OK.
b) 3.13m/s is OK.
c) No. You have used a formula for freefall (acceleration = g); the question says his feet are in contact with the ground so his acceleration is NOT g.
The 0.4m must refer to distance moved rising from a crouched position to just taking-off,
Initial velocity, u = 0 (crouched)
final velocity, v = 3.13m/s (taking off)
distance, d = 0.4m
d = (u+v)t/2 (same as saying distance = (avge. vel) x time)
0.4 =(0+3.13) t / 2
t = 0.8/3.13
= 0.256s
d) ) No. I have no idea what formula you are using. The average power will be
P = (total energy supplied) / (time.taken to supply it)
= 400/0.256 = 1560W (to 3 sig figs - no more)
b) 3.13m/s is OK.
c) No. You have used a formula for freefall (acceleration = g); the question says his feet are in contact with the ground so his acceleration is NOT g.
The 0.4m must refer to distance moved rising from a crouched position to just taking-off,
Initial velocity, u = 0 (crouched)
final velocity, v = 3.13m/s (taking off)
distance, d = 0.4m
d = (u+v)t/2 (same as saying distance = (avge. vel) x time)
0.4 =(0+3.13) t / 2
t = 0.8/3.13
= 0.256s
d) ) No. I have no idea what formula you are using. The average power will be
P = (total energy supplied) / (time.taken to supply it)
= 400/0.256 = 1560W (to 3 sig figs - no more)
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I disagree with Steve4Ph's answers. Why was this question resolved so quickly? AC, I suggest you re-pose the question. I have my solution but there's not enough space here to elaborate. My answers are b) 1.4 m/s; c) 0.57 s, d) 563 W. We all agree on (a).
U.N. Owen
U.N. Owen
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