Question

Please, someone help me I'm begging yall 1. it's estimated that 1 kg of body fat will provide 3.8 * 10^7 J of energy. A 67 kg mountain climber decides to climb a mountain 3500 m high. a) How much work does the climber do against gravity in climbing to the top of the mountain? b) if the body's effiency in converting energy stored as fat to mechanical engery is 25% determine the amount of fat the climber will use up in providing the energy required to work against the force

Answer 1

Answer:

0.24 kg used up

Explanation:

He has a mass of 67 kg

The gravitational constant is 9.81 m/s^2

The distance upward is 3500 m

W = m*g*h

W = 67 * 9.81 * 3500

Work = 2,300,445 Joules

Work = 2300 kj

work = 2.30 * 10^6 joules in scientific notation.

Part B

He needs 4 times this amount to climb the mountain because the body is only 25% efficient in converting energy.

4*2.30 * 10^6 = 9.20 * 10^6 Joules of energy are therefore required.

The total amount in a kg of fat = 3.8 * 10^7 joules

x kg of fat is needed to provide 9.20.*10^6 joules

1 kg / (3.8 * 10^7 J ) = x kg / (9.20 * 10^6 J)

9.20 * 10 ^6 * 1 = 3.8 * 10^7 *x

9.20 * 10 ^6 / 3.8 * 10^7 = x

x = 0.24 kg of fat are needed