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Chapter 14: Problem 14

An \(83-\mathrm{kg}\) man eats a banana of energy content $1.00 \times 10^{2} \mathrm{kcal} .$ If all of the energy from the banana is converted into kinetic energy of the man, how fast is he moving, assuming he starts from rest?

Short Answer

Expert verified
Answer: Approximately 31.43 m/s.

Step by step solution

01

Write down the given information

Mass of the man (m) is given as 83 kg. Energy content of the banana (E) is given as 1.00 × 10^2 kcal.
02

Convert energy content from kcal to Joules

To work with SI units, we need to convert the energy content of the banana from kcal to Joules (J). 1 kcal = 4184 J So, E = 1.00 × 10^2 kcal × 4184 J/kcal = 4.184 × 10^5 J
03

Write down the formula for kinetic energy

The formula for kinetic energy (K) when an object starts from rest is given by: K = 0.5 * m * v^2 Where m is the mass and v is the final speed.
04

Set the energy content equal to the kinetic energy and solve for speed (v)

According to our problem, all the energy from the banana will be converted into kinetic energy of the man. 4.184 × 10^5 J = 0.5 * 83 kg * v^2 Now, we need to solve for v: v^2 = (4.184 × 10^5 J) / (0.5 * 83 kg) v^2 ≈ 10096.38 v = sqrt(10096.38) v ≈ 31.43 m/s
05

Conclusion

If all the energy from the banana is converted into kinetic energy of the man, he will be moving at a speed of approximately 31.43 m/s.

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Most popular questions from this chapter

If the total power per unit area from the Sun incident on a horizontal leaf is \(9.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2},\) and we assume that \(70.0 \%\) of this energy goes into heating the leaf, what would be the rate of temperature rise of the leaf? The specific heat of the leaf is $3.70 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right),$ the leaf's area is \(5.00 \times 10^{-3} \mathrm{m}^{2},\) and its mass is $0.500 \mathrm{g}$.
A 10.0 -g iron bullet with a speed of $4.00 \times 10^{2} \mathrm{m} / \mathrm{s}\( and a temperature of \)20.0^{\circ} \mathrm{C}$ is stopped in a \(0.500-\mathrm{kg}\) block of wood, also at \(20.0^{\circ} \mathrm{C} .\) (a) At first all of the bullet's kinetic energy goes into the internal energy of the bullet. Calculate the temperature increase of the bullet. (b) After a short time the bullet and the block come to the same temperature \(T\). Calculate \(T\), assuming no heat is lost to the environment.
A 64 -kg sky diver jumped out of an airplane at an altitude of $0.90 \mathrm{km} .$ She opened her parachute after a while and eventually landed on the ground with a speed of \(5.8 \mathrm{m} / \mathrm{s} .\) How much energy was dissipated by air resistance during the jump?
If \(125.6 \mathrm{kJ}\) of heat are supplied to \(5.00 \times 10^{2} \mathrm{g}\) of water at \(22^{\circ} \mathrm{C},\) what is the final temperature of the water?
Convert \(1.00 \mathrm{kJ}\) to kilowatt-hours \((\mathrm{kWh})\).
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