Question

Physics Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

A) The first one is simply potential energy  mgh. Great you got it already.

b)  1 kg of body fat would provide 3.8 x 10^7 J of energy.
 
The conversion rate to mechanical work is 25% = 0.25

1 kg would therefore provide (3.8 * 0.25 * 10^7)  = 0.95 * 10 ^ 7 J of mechanical work.

1 kg of fat provided 9.5 * 10^6 J of mechanical work.

x kg  of fat would be needed for 2.4 * 10^ 6 J of mechanical work.

x =    (2.4 * 10^ 6 J) * 1 kg / (9.5 * 10^6 J)    Use your calculator.

x = 0.2526 kg of fat would be needed.

That's the amount. Cheers.

Answer 2

Good work on solving part a).
b) may look complicated, but it's not too bad.

It says that the body is 25% efficient in converting fat to mechanical energy.
In other words, only 25% of the energy we get from our stored fat shows up
in the physical, mechanical moving around that we do.  (The rest becomes
heat, which dissipates into the environment as we keep our bodies warm,
breathe hot air out,and perspire.)

You already know how much mechanical energy the climber needed to lift
himself to the top of the mountain... 2.4x10⁶ joules.
That's 25% of what he needs to convert in order to accomplish the climb.
He needs to pull 4 times as much energy out of fat.

-- Fat energy required = 4 x (2.4 x 10⁶) = 9.6 x 10⁶ joules.

-- Amount stored in 1kg of fat = 3.8 x 10⁷ joules

-- Portion of a kilogram he needs to use =   (9.6 x 10⁶) / (3.8 x 10⁷)

Note:
That much of a kilogram weighs about 8.9 ounces ... which shows why it's so
hard to lose weight with physical exercise alone.  It also helps you appreciate
that fat is much more efficient at storing energy than batteries are ... that one
kilogram of fat stores the amount of energy used by a 100-watt light bulb, to
burn for 105 hours (more than 4-1/2 days ! ! !)