Question

How many grams of butter, which has a usable energy content

of6.0 Cal/g (= 6000 cal/g), would be equivalent to the change
ingravitational potential energy of a 60 kg man who climbs 6.70
km?Assume that the average gfor the climb is 9.80 m/s2.

Answer 1

`m_(butter) = 156.93 grams`

Step-by-step explanation:

Given data;

energy content = 6.0 Cal/g = 6000 cal/g

mass of man 60 kg

climb height = 6.70 km

g = 9.80 m/s^2

we know that

potential energy = mgh

`E_t = mU`

wehre U - usable energy constant

equating both energy

`60 * 9.8 * 6.70* 10^3 = m_(butter) 6000* (4.184 j)/(cal)`

`m_(butter) = 156.93 grams`