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Chapter 8: Problem 67

Taking a leap of faith, a bungee jumper steps off a platform and falls until the cord brings her to rest. Suppose you analyze this system by choosing \(y=0\) at the platform level, and your friend chooses \(y=0\) at ground level. (a) Is the jumper's initial potential energy in your calculation greater than, less than, or equal to the same quantity in your friend's calculation? Explain. (b) Is the change in the jumper's potential energy in your calculation greater than, less than, or equal to the same quantity in your friend's calculation? Explain.

Short Answer

Expert verified
(a) Less than your friend's calculation. (b) Equal in both calculations.

Step by step solution

01

Understand the Concept of Potential Energy

Potential energy depends on the reference point where we choose to identify zero gravitational potential energy. When you calculate gravitational potential energy, it is given by the formula, \( U = mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is the height above your chosen reference point.
02

Identify the Reference Points

In your calculation, the reference point for \( y = 0 \) is the platform level. In your friend's calculation, the reference point for \( y = 0 \) is at ground level. Potential energy will differ based on these reference points.
03

Analyze Initial Potential Energy

(a) The jumper鈥檚 initial potential energy in your calculation uses the platform (\( y = 0 \)) as the reference point, hence it's \( U = 0 \). In your friend's calculation, the initial potential energy uses \( h \) as the distance from the ground to the platform, thus it's \( U = mgh \). Therefore, the jumper's initial potential energy in your calculation is less than your friend's.
04

Analyze Change in Potential Energy

(b) The change in potential energy is defined by the difference between initial and final states. Since both you and your friend calculate using the same physical scenario, just different reference points, the change in potential energy depends on the same physical displacement and mass, independent of the starting zero point. Thus, your calculated change in potential energy is equal to the same quantity in your friend's calculation.

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Key Concepts

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

Reference Point
In physics, the reference point is crucial when calculating different quantities like potential energy. Here, it's a position where you define the potential energy to be zero. Choosing a reference point may seem arbitrary, but it significantly impacts calculations.

Let's consider a bungee jumper. If you set the reference point at the platform where the jump begins, any height above will be positive, and any height below will be negative. On the other hand, if you pick the ground as the reference point, all heights are non-negative.
  • Different reference points can result in different initial potential energy values.
  • The overall physics of the situation remains the same since physics calculations depend on changes rather than absolute values.
Therefore, while the choice of reference point affects initial potential energy calculations, it does not affect the changes observed in a system.
Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses due to its position relative to Earth. It depends on three things:
  • Mass of the object (\(m\))
  • Gravitational acceleration (\(g\)), roughly \(9.8 \text{ m/s}^2\) on Earth
  • Height (\(h\)) above the chosen reference point
The calculation is simple: \(U = mgh\).

When the bungee jumper is on the platform, her gravitational potential energy depends on the reference point:
  • If the reference point is the platform (\(y = 0\)), \(U = 0\)
  • If set at ground level, the potential energy is maximized because \(U = mgh\)
Thus, understanding gravitational potential energy helps explain why the jumper's potential energy differs when different reference points are used.
Bungee Jumping
Bungee jumping is a thrilling practical application of physics concepts. It involves converting potential energy into kinetic energy. When a bungee jumper steps off the platform, the gravitational potential energy begins to decrease, converting into kinetic energy as they fall.

Key points during the jump include:
  • The highest potential energy is at the starting point unless the reference is set there (then it's zero).
  • As they fall, potential energy decreases while kinetic energy increases.
  • The bungee cord stretches to absorb and dissipate energy.
  • At the lowest point, the potential energy is zero if the reference is on the ground and maximum kinetic energy is reached as the jumper momentarily stops.
Bungee jumping vividly demonstrates energy conservation and transformation principles in a real-world context.
Physics Problem Solving
Physics problem-solving often involves choosing a correct and strategic approach to analyze situations. When calculating quantities like potential energy, identify the system and reference points clearly; this impacts your calculations.

For instance, in a bungee jumping problem, important steps include:
  • Identifying the chosen reference point for zero potential energy.
  • Calculating potential energy at various points using \(U = mgh\) based on that reference.
  • Understanding that change in potential energy remains consistent regardless of the reference point.
  • Applying energy conservation principles to understand energy transformation.
Effective problem-solving relies on understanding not just calculations, but the underlying principles. Practice these methods regularly, and they will become second nature in analyzing physics problems, making the solutions more intuitive.

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Most popular questions from this chapter

The driver of a \(1300-\mathrm{kg}\) car moving at \(17 \mathrm{m} / \mathrm{s}\) brakes quickly to \(11 \mathrm{m} / \mathrm{s}\) when he spots a local garage sale. (a) Find the change in the car's kinetic energy. (b) Explain where the "missing" kinetic energy has gone.

A sled slides without friction down a small, ice-covered hill. If the sled starts from rest at the top of the hill, its speed at the bottom is \(7.50 \mathrm{m} / \mathrm{s}\). (a) On a second run, the sled starts with a speed of \(1.50 \mathrm{m} / \mathrm{s}\) at the top. When it reaches the bottom of the hill, is its speed \(9.00 \mathrm{m} / \mathrm{s},\) more than \(9.00 \mathrm{m} / \mathrm{s},\) or less than \(9.00 \mathrm{m} / \mathrm{s}\) ? Explain. (b) Find the speed of the sled at the bottom of the hill after the second run.

Predict/Explain When a ball of mass \(m\) is dropped from rest from a height \(h,\) its speed just before landing is \(v .\) Now, suppose a second ball of mass \(4 m\) is dropped from rest from a height \(h / 4 .\) (a) Just before ball 2 lands, is its speed \(4 v, 2 v, v, v / 2\) or \(v / 4\) ? (b) Choose the best explanation from among the following: I. The factors of 4 cancel; therefore, the landing speed is the same. II. The two balls land with the same kinetic energy; therefore, the ball of mass \(4 m\) has the speed \(v / 2\) III. Reducing the height by a factor of 4 reduces the speed by a factor of 4.

A player passes a \(0.600-\mathrm{kg}\) basketball downcourt for a fast break. The ball leaves the player's hands with a speed of \(8.30 \mathrm{m} / \mathrm{s}\) and slows down to \(7.10 \mathrm{m} / \mathrm{s}\) at its highest point. (a) lgnoring air resistance, how high above the release point is the ball when it is at its maximum height? (b) How would doubling the ball's mass affect the result in part (a)? Explain.

At the local playground a child on a swing has a speed of \(2.02 \mathrm{m} / \mathrm{s}\) when the swing is at its lowest point. (a) To what maximum vertical height does the child rise, assuming he sits still and "coasts"? Ignore air resistance. (b) How do your results change if the initial speed of the child is halved?
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