The problem states: You curl a 5.5kg barbell that is hanging straight down in yo

Question

The problem states: You curl a 5.5kg barbell that is hanging straight down in your hand up to your shoulder.

Determine the work the hand does in lifting the barbell. The barbell is lifting up about 1.0m. The length of your arm.

W = 54 J solved.

Determine the average mechanical power of the lifting process. Suppose it takes 0.3s to lift it up.

P = W/t = F*d/t = F*v, but for this case, P = W/t.

Answer = 179.85 W or (J/s)

Now,  if the body is 20% efficient at converting mechanical energy into chemical energy, how many times would you need to lift the barbell in order to burn enough calories to use absorbed by a 300 food calorie dish?  1 food cal = 1 kilo cal = 4186 J

How can I solve this? I have 4186 J * 300 = 1255800 J.

Now what?

Explanation / Answer

let's assume n times

so n*54*.2=1255800

n=116277.7=116278(approx.)

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