A gymnast has a mass of 64 kg. a) A gymnast is in a handstand position on a 2.60
ID: 1395842 • Letter: A
Question
A gymnast has a mass of 64 kg. a) A gymnast is in a handstand position on a 2.60 m high bar. The gymnast?s center of mass is located 0.85 m above the bar. If the gymnast is performing a giant swing (full revolution around the bar), what is the gymnast?s velocity at the lowest point in the swing? b) During an awkward dismount, the gymnast slides 0.21 m during her landing. If the coefficient of friction between the gymnast?s feet and the mat is 0.72, what was her horizontal velocity when she initially landed? c) During a successful dismount, the mat deflects 0.038 m downward. If the stiffness of the mat is 320,000 N/m, what was the downward vertical velocity at the gymnast?s initial contact with the mat? Assume changes in potential energy are negligible. d) If the gymnast elevates her center of mass 0.85 m up to the high bar in 1.39 s, how much power does she generate? e) In part a), what is the gymnast?s total mechanical energy at the bottom of the swing? A. the total mechanical energy is negative B. the total mechanical energy is zero C. the total mechanical energy is the same as at the top of the swing D. the total mechanical energy is greater than at the top of the swingExplanation / Answer
a) potential energy of the gymnast in hand stand position PE1= m*g*h
Kinetic energy of the gymnast = KE1 = 0
total energu TE1 = m*g*h
potential energy of the gymnast at the lower position = -m*g*h
KE2 = 0.5*m*v^2
total energy at lower position TE2 = 0.5*m*v^2 - m*g*h
from energy conservation
TE2 = TE1
0.5*m*v^2 - m*g*h = m*g*h
h = 0.85 cm
0.5*m*v^2 = 2*m*g*h
v = sqrt(4*g*h) = 5.8 m/s <--------answer
option C
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b)
acceleration = ax = -u*g = -0.72*9.8 = -7.1 m/s^2
initial velocity = vi
the gymnast is coming to rest so,final velocity = vf = 0
from equations of motion
vf^2 - vi^2 = 2*ax*x
0^2 - vi^2 = -2*7.1*0.21
vi = 1.7 m/s<-answer
option B
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c)
potential energy stored in the mat = U = 0.5*k*y^2 = 0.5*320000*0.038^2 = 231.04 J
kinetic energy of the gymnast before coming in to contact = KE = 0.5*m*v^2
from energy conservation
KE = PE
0.5*m*v^2 = 231.04
v = 2.7 m/s <----answer
option B
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d) potential energy of the gymnast = PE = m*g*h = 64*9.8*0.85) = 533.12 J
power = 533.12/1.39 = 384 W
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e)
as there are no frictional force as the gymast is swinging there is no loss of energy
the total mechanical energy is the same as the top of the swing
option C
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