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The food calorie, equal to \(4186 \mathrm{~J},\) is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit- and-cereal bar contains 140 food calories per bar. (a) If a \(65 \mathrm{~kg}\) hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy? (b) If, as is typical, only \(20 \%\) of the food calories go into mechanical energy, what would be the answer to part (a)?

Short Answer

Expert verified
(a) The hiker must climb approximately 922.57 meters. (b) With only 20% energy conversion, he must climb approximately 184.51 meters.

Step by step solution

01

Calculate Energy from the Food Bar

Convert the energy from the food bar to joules. Since 1 food calorie is equivalent to 4186 joules, for a 140 calorie bar, calculate the energy in joules:\[E = 140 \text{ calories} \times 4186 \text{ J/calorie} = 586,040 \text{ J}\]
02

Calculate Height for Full Energy Conversion

To find the height the hiker can climb, assuming 100% energy conversion to mechanical energy, use the formula for gravitational potential energy \(E = mgh\), where \(m = 65 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(E = 586,040 \text{ J}\). Solve for \(h\):\[586,040 = 65 \times 9.8 \times h \h = \frac{586,040}{65 \times 9.8} \approx 922.57 \text{ m}\]
03

Calculate Height with 20% Energy Conversion to Mechanical Energy

If only 20% of the energy goes into mechanical energy, calculate the usable energy:\[E_{usable} = 0.2 \times 586,040 \text{ J} = 117,208 \text{ J}\]Now, using the same potential energy formula but with \(E_{usable}\), solve for \(h\):\[117,208 = 65 \times 9.8 \times h \h = \frac{117,208}{65 \times 9.8} \approx 184.51 \text{ m}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Potential Energy
Gravitational potential energy is an important concept in physics that describes the energy stored in an object due to its position in a gravitational field. This energy results from an object's height above a reference point, usually the ground. The more elevated the object, the greater the gravitational potential energy it possesses. This is because it has more potential for motion under the influence of gravity.

The formula for gravitational potential energy is expressed as:
  • \( E = mgh \)
Here, \( E \) represents the gravitational potential energy in joules, \( m \) is the mass of the object in kilograms, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth), and \( h \) is the height in meters.

This concept is particularly useful in problems involving changes in altitude where energy is transferred or transformed, as in the case of a hiker climbing a mountain to "work off" energy from a food caloric intake.
Mechanical Energy
Mechanical energy is the sum of potential energy and kinetic energy present in the components of a mechanical system. It is the energy associated with the motion and position of an object. In other words, mechanical energy is the energy acquired by the object due to its motion (kinetic energy) and its position (potential energy).

In our exercise scenario, we consider the hiker's mechanical energy when converting food-derived energy into potential energy while climbing a mountain. Typically, not all of the energy consumed from food is converted into mechanical energy. Often, only a portion, such as 20%, effectively contributes to work done against gravity.

By focusing on potential energy in this exercise, we assume that the mechanical energy applied by the hiker is fully utilized for elevation gain, whether 100% or 20% of the caloric energy is converted to mechanical energy, highlighting the energy conversion efficiency in biological systems.
Food Calories
Food calories are units of energy that quantify the amount of energy we gain from food when metabolized. Scientifically, a food calorie found on nutritional labels is equivalent to 4186 joules. This energy is vital for various bodily functions and can be harnessed for mechanical tasks, such as climbing or running.

In practical terms, food provides the energy necessary for our muscles and organs to function. The number of calories in a food item tells us about the potential energy it can release, thereby informing our dietary and exercise decisions.

Understanding food calories also allows us to calculate energy use and balance, as seen in our exercise involving the hiker. Here, we translate the energy (140 food calories) from a snack bar into joules, which gives a clearer picture of how much work that energy can potentially support.
Potential Energy Calculation
In our problem-solving scenario, calculating potential energy involves identifying how far a hiker can climb after consuming specific food energy. This begins with converting food calories to joules, acknowledging the caloric content of food as a measure of stored energy.

Once we have the energy in joules, we use the expression for gravitational potential energy, \( E = mgh \), to find the maximum altitude the hiker could theoretically reach by assuming the energy is only used for climbing. This calculation provides a direct way to relate consumed calories to physical activity levels.

For the exercise, we calculated two scenarios: one with full energy conversion and another where only 20% of energy is converted to mechanical energy. These different outcomes showcase how energy conversion efficiencies dramatically influence achievable potential heights during physical activities. By performing these calculations, we understand how caloric intake translates into mechanical energy that moves our bodies against gravitational forces.

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Most popular questions from this chapter

A \(1.50 \mathrm{~kg}\) book is sliding along a rough horizontal surface. At point \(A\) it is moving at \(3.21 \mathrm{~m} / \mathrm{s},\) and at point \(B\) it has slowed to \(1.25 \mathrm{~m} / \mathrm{s}\). (a) How much work was done on the book between \(A\) and \(B ?\) (b) If \(-0.750 \mathrm{~J}\) of work is done on the book from \(B\) to \(C\) how fast is it moving at point \(C ?\) (c) How fast would it be moving at \(C\) if \(+0.750 \mathrm{~J}\) of work were done on it from \(B\) to \(C ?\)

An elevator has mass \(600 \mathrm{~kg}\), not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of \(20.0 \mathrm{~m}\) (five floors) in \(16.0 \mathrm{~s}\), and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass \(65.0 \mathrm{~kg}\).

II A slingshot obeying Hooke's law is used to launch pebbles vertically into the air. You observe that if you pull a pebble back \(20.0 \mathrm{~cm}\) against the elastic band, the pebble goes \(6.0 \mathrm{~m}\) high. (a) Assuming that air drag is negligible, how high will the pebble go if you pull it back \(40.0 \mathrm{~cm}\) instead? (b) How far must you pull it back so it will reach \(12.0 \mathrm{~m} ?\) (c) If you pull a pebble that is twice as heavy back \(20.0 \mathrm{~cm},\) how high will it \(\mathrm{go} ?\)

A spring is \(17.0 \mathrm{~cm}\) long when it is lying on a table. One end is then attached to a hook and the other end is pulled by a force that increases to \(25.0 \mathrm{~N}\), causing the spring to stretch to a length of \(19.2 \mathrm{~cm}\). (a) What is the force constant of this spring? (b) How much work was required to stretch the spring from \(17.0 \mathrm{~cm}\) to \(19.2 \mathrm{~cm} ?\) (c) How long will the spring be if the \(25 \mathrm{~N}\) force is replaced by a \(50 \mathrm{~N}\) force?

Your friend (mass \(65.0 \mathrm{~kg}\) ) is standing on the ice in the middle of a frozen pond. There is very little friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for \(3.00 \mathrm{~s}\) and accelerate your friend from rest to a speed of \(6.00 \mathrm{~m} / \mathrm{s}\) while you remain at rest. What is the average power supplied by the force you applied?

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