Professor Goetz took a heavy marine 100 foot, 0.75 lb per foot rope and through it through his office window. One end of the rope touches the ground. Professor Goetz missed his lunch and can only exert 300kcal (the equivalent of a healthy sandwich). He realized he does not know if he has enough energy to lift the rope back into his office. a) How much energy in kcalories is needed to lift the rope half way from the ground. b) Calculate whether he is capable of lifting the rope all the way up. c) A person who eat a candy gains extra 50kcal. If this energy is used to lift the rope, how many feet will it be lifted from the ground.

Question

Cassie Jefferson

Physics

12 November, 04:46

Professor Goetz took a heavy marine 100 foot, 0.75 lb per foot rope and through it through his office window. One end of the rope touches the ground. Professor Goetz missed his lunch and can only exert 300kcal (the equivalent of a healthy sandwich). He realized he does not know if he has enough energy to lift the rope back into his office. a) How much energy in kcalories is needed to lift the rope half way from the ground. b) Calculate whether he is capable of lifting the rope all the way up. c) A person who eat a candy gains extra 50kcal. If this energy is used to lift the rope, how many feet will it be lifted from the ground.

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Answers (1)

Know the Answer?

Stewart · Answer 1

a. E = 11.90876 kcal

b. E = 23.81753 k cal

c. d = 209.93 ft

Explanation:

Knowing the relation between the energy and the food to realize how may energy do you need in each situation

a.

h = 100 ft / 2 = 50 ft

We = 0.75 lb * 100 * 9.8 = 735 J

E = We * h = 735 J * 50 ft = 36750 J

E to Kcal = 36750 J ⇒ 11.90876 k cal

E = 11.90876 kcal

b.

E = 735 * 100 ft = 73500 J

E to Kcal 73500 J ⇒ 23.81753 k cal

E = 23.81753 k cal

c.

Now to determine the distance with a candy of 50 kcal

d = 100 ft ⇒ 23. 81753 k cal

d ⇒ 50 k cal

d = (100 * 50) / 23.81753

d = 209.93 ft