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The maximum height a typical human can jump from a crouched start is about 60 cm. By how much does the gravitational potential energy increase for a 72 kg person in such a jump? Where does this energy come from?

Short Answer

Expert verified
The gravitational potential energy increases by about 423.36 J. This energy comes from the work done by the person's muscles during the jump, converting energy from food into kinetic and potential energy.

Step by step solution

01

Converting height into meters

The height of the jump is given in centimeters. Before solving, convert it to meters. Since 1 meter is equal to 100 centimeters, we do this by dividing by 100: \[ 60 \, cm \div 100 = 0.6 \, m \]
02

Find Potential Energy Increase

Use the potential energy formula \[ PE = mgh \]. Substitute the known values into this equation \[ PE = 72 \, kg \times 9.8 \, m/s² \times 0.6 \, m \]. By multiplying these numbers, the increased gravitational potential energy can be found.
03

Explaining the Source of Energy

The energy comes from the work done by the person’s muscles during the jump. Energy from the food they eat is changed in the body into kinetic and potential energy. As they jump, kinetic energy is transferred into gravitational potential energy.

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Key Concepts

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

Potential Energy Formula
Understanding the potential energy formula is crucial when calculating the gravitational potential energy of an object in a physics problem. Gravitational potential energy (PE) is the energy an object possesses due to its position in a gravitational field. The standard formula for gravitational potential energy is given by:
\[ PE = mgh \]
where:
  • \( m \) is the mass of the object in kilograms,
  • \( g \) is the acceleration due to gravity (9.8 m/s²\ton Earth),
  • \( h \) is the height above the reference point in meters.
To solve the example problem, we plug in the values for the mass of the person, the acceleration due to gravity, and the height they have jumped to calculate the increase in gravitational potential energy. It's essential to ensure that the unit of the height is converted to meters, as consistent units are fundamental for accurate calculations. The mass of the person is given as 72 kg, and the height of the jump is 60 cm, which we convert to 0.6 m. By substituting these values into the formula, we can find that the energy required for a 72 kg person to reach a height of 0.6 m is:
\[ PE = 72 kg \times 9.8 m/s² \times 0.6 m \]
Work-Energy Principle
The work-energy principle is a fundamental concept in physics that relates the work done on an object to the change in its energy. When work is performed on an object, it results in a transfer and transformation of energy. Work done against gravitational forces results in an increase in gravitational potential energy of an object.
In our scenario, when a person jumps, their muscles do work to propel them upward. This work is equivalent to the increase in gravitational potential energy as they reach the maximum height of their jump. The formula for work (W\ton) in terms of energy is expressed as:
\[ W = \text{Change in energy} = PE_{final} - PE_{initial} \]
If we consider the ground level as the point where the potential energy is zero (\( PE_{initial} = 0 \)), then the work done by the muscles is equal to the final potential energy at the peak of the jump (\( PE_{final} \)). This demonstrates the conversion of kinetic energy generated by the muscles into potential energy at the jump's apex.
Unit Conversion
Unit conversion is a critical step in solving many physics problems, ensuring quantities are measured in compatible units. It involves transforming a measure expressed in one set of units into another, maintaining the same quantity. In the context of our example, we need to convert the jump's height from centimeters to meters before we can use it in the potential energy formula.
This is because the standard unit for length in the International System of Units (SI) is meters (m), while the formula for gravitational potential energy requires that all units are consistent. Here's how we perform the conversion:
\[ 60 \text{cm} \tdiv 100 = 0.6 m \]
There are 100 centimeters in 1 meter, so dividing by 100 accomplishes the conversion. It's always helpful to remember some common conversions, such as 1 km = 1000 m or 1 inch = 2.54 cm, as they are frequently used. Being proficient at unit conversion can prevent errors in calculations and is a valuable skill in not only physics but in various scientific and engineering fields.

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Most popular questions from this chapter

A 90.0 kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 m sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?

A block with mass \(0.400 \mathrm{~kg}\) is on a horizontal frictionless surface and is attached to a horizontal compressed spring that has force constant \(k=200 \mathrm{~N} / \mathrm{m} .\) The other end of the spring is attached to a wall. The block is released, and it moves back and forth on the end of the spring. During this motion the block has speed \(3.00 \mathrm{~m} / \mathrm{s}\) when the spring is stretched \(0.160 \mathrm{~m}\). (a) During the motion of the block, what is its maximum speed? (b) During the block's motion, what is the maximum distance the spring is compressed from its equilibrium position? (c) When the spring has its maximum compression, what is the speed of the block and what is the magnitude of the acceleration of the block?

A small rock with mass 0.20 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with radius R = 0.50 m (Fig. E7.9). Assume that the size of the rock is small com-pared to R, so that the rock can be treated as a particle, and assume that the rock slides rather than rolls. The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl has magnitude 0.22 J. (a) Between points A and B, how much work is done on the rock by (i) the normal force and (ii) gravity? (b) What is the speed of the rock as it reaches point B? (c) Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? Explain. (d) Just as the rock reaches point B, what is the normal force on it due to the bottom of the bowl?

The food calorie, equal to \(4186 \mathrm{~J},\) is a measure of how much energy is released when the body metabolizes food. A certain fruit-and-cereal bar contains 140 food calories. (a) If a 65 kg hiker eats one bar, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes 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)? (Note: In this and all other problems, we are assuming that \(100 \%\) of the food calories that are eaten are absorbed and used by the body. This is not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the body eliminates the rest. Metabolic efficiency varies considerably from person to person.

\(\mathrm{A} 1500 \mathrm{~kg}\) rocket is to be launched with an initial upward speed of \(50.0 \mathrm{~m} / \mathrm{s} .\) In order to assist its engines, the engineers will start it from rest on a ramp that rises \(53^{\circ}\) above the horizontal (Fig. \(\mathbf{P 7 . 5 0}\) ). At the bottom, the ramp turns upward and launches the rocket vertically. The engines provide a constant forward thrust of \(2000 \mathrm{~N}\), and friction with the ramp surface is a constant \(500 \mathrm{~N}\). How far from the base of the ramp should the \text { rocket start, as measured along the surface of the ramp?

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