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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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