Question

A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 43 kg barbell straight up from chest height to full arm extension, adistance of 0.54m . How much work does the weightlifter do to lift the barbell one time? If the weightlifter does 17 repetitions a day, what total energy does sheexpend on lifting, assuming a typical efficiency for energy use by the body? How many 400 Calorie donuts can she eat a day to supply that energy?

Answer 1

A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 43kg barbell straight up from chest height to full armextension, a distance of 0.54m.

(A) How much work does the weightlifter do to lift the barbell one time?
Work = force * distance
Weight = 43 * 9.8
Work = 43* 9.8 * 0.54 = 227.56 N-m

(B) If the weightlifter does 17 repetitions a day, what total energy does she expend on lifting? Assume 17% efficiency.
Work = Energy
Energy =227.56 J /rep * 17 rep = 3868.45 J if 100% efficiency
Since she is only 17% efficient, she has to expend 5390/0.17 J = 22755.6 J of energy to do 5390 J of work.

(C) How many 400Calorie donuts can she eat a day to supply that energy?
4.18 J = 1 calorie
22755.6 J ÷ 4.18 J/cal = 5443.9 calories ÷ 400cal /donut = 13.6 donuts

Answer 2

Given
Mass of barbell (m) = 40 kg
Distance S = 0.55 m
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( A).Work does the weightlifter do to lift the barbell one time W = mgS
= 215.6 J
= 210 J
(B).No.of repetations per day (n) = 17
Total work done to lift the barbell n times W ' = nW
= 3665.2 J
efficiency (e )= 25 %
= 0.25
? Total energy does she expend on lifting E = W ' / e
= 14660.8 J
= 14000 J
(c) Energy of each donut E ' = 480 cal
require energy to produce one joule of energy E' = 480 / 4.2
no .of calories dounut to eat = (480 /14000* 4.2 )
=0.007.

Answer 3

Part A

Work Done = Change in Potential Energy

= mgh

= 41*9.8*0.45

= 180.81 J

Part B

A typical efficiency for energy is a Standard term menas 25% of Efficiency

Therefore

Work Done = No. of repetitions*mgh/Efficiency

= (23*41*9.8*0.45)/(0.25)

= 16634.52 J

Part C

Let the no. of Donuts = N

Therefore

N*Energy in one Donut = Energy required

N*500*4184 = 16634.52

N = 0.00795

= 0.795% of the one Donut