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For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the efficiency of the human body is 25%, and that he lifts the barbell at a constant speed. Show all work and include proper unit for your final answer.
a) In applying the energy equation (ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W) to the system consisting of the earth, the barbell, and the athlete,
1. Which terms (if any) are positive?
2. Which terms (if any) are negative?
3. Which terms (if any) are zero?
b) Determine the energy output by the athlete in SI unit.
c) Determine his metabolic power in SI unit.
d) Another day he performs the same task in 1.2 s.
1. Is the metabolic energy that he expends more, less, or the same?
2. Is his metabolic power more, less, or the same?

Respuesta :

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

batolisis

A) Applying the energy equation

  1. The positive terms is :   ΔUg
  2. The negative terms is :  ΔEth
  3. The zero term are :  ΔK  and ΔUs

B) The energy output by the athlete is ; 800 Joules

C) The metabolic power is : 2000 w

D) When he performs the task in 1.2 s

  1. The metabolic energy he expends is : the same
  2. His metabolic power is :  more

Given data :

Weight of barbell = 400 N

Height = 2.0 m

Time = 1.6 secs

efficiency of the human body = 25%

Speed = constant

A) From the energy equation the ΔK is zero because the athlete is lifting the barbell at a constant speed. ΔUg is positive because as the weight is lifted its  potential energy increases.  ΔEth ( change in energy of earth ) is negative because it exerts a force in opposite direction to displacement

B)  Determine the energy output of the athlete

weight of barbell * Height  = 400 * 2 = 800 J

C) Determine the metabolic power

Metabolic power = energy output / Time

where ; energy output = 4 * 800 = 3200

∴ Metabolic power = 3200 / 1.6

                                = 2000 w

D) When performs same task at 1.2 s

The metabolic energy he expends is  the same  and His metabolic power is  more

Hence we can conclude that the answers to your questions are as listed above

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