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Chapter 6: Problem 30

Juggles and Bangles are clowns. Juggles stands on one end of a teeter-totter at rest on the ground. Bangles jumps off a platform \(2.5 \mathrm{m}\) above the ground and lands on the other end of the teeter-totter, launching Juggles into the air. Juggles rises to a height of \(3.3 \mathrm{m}\) above the ground, at which point he has the same amount of gravitational potential energy as Bangles had before he jumped, assuming both potential energies are measured using the ground as the reference level. Bangles' mass is \(86 \mathrm{kg}\). What is Juggles' mass?

Short Answer

Expert verified
Juggles' mass is approximately 65 kg.

Step by step solution

01

Write the formula for gravitational potential energy

Gravitational potential energy is given by the formula: \[ U = mgh \] where \( U \) is the potential energy (in joules), \( m \) is the mass (in kilograms), \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( h \) is the height (in meters).
02

Calculate Bangles' initial potential energy

Bangles' potential energy before jumping can be calculated as:\[ U_B = m_B \, g \, h_B = 86 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 2.5 \, \text{m} \]Solving this gives:\[ U_B = 2,110.35 \, \text{J} \]
03

Define the condition of equal potential energy for Juggles

According to the problem, Juggles' potential energy when he reaches a height of \(3.3 \, \text{m}\) is equal to Bangles' initial potential energy. Hence, we have:\[ U_J = U_B \] where \( U_J \) is Juggles' potential energy and \( U_B = 2,110.35 \, \text{J} \).
04

Calculate Juggles' mass

Using the formula for gravitational potential energy, we substitute for Juggles:\[ U_J = m_J \, g \, h_J \]Substituting the known values:\[ 2,110.35 \, \text{J} = m_J \times 9.81 \, \text{m/s}^2 \times 3.3 \, \text{m} \]Solving for \( m_J \), we get:\[ m_J = \frac{2,110.35}{9.81 \times 3.3} \approx 65.0 \, \text{kg} \]

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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 a type of energy that objects possess because of their position in a gravitational field. It can be easily calculated using the formula: \[ U = mgh \]where:
  • \( U \) is the gravitational potential energy,
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity, typically \( 9.81 \text{ m/s}^2 \) on Earth,
  • \( h \) is the height above a reference point, often the ground.
The energy is measured in joules. When a mass is raised, its potential energy increases. Conversely, when it falls, energy is reduced.

Consider Bangles the clown, before he jumps鈥攕tanding at a height of \( 2.5 \text{ m} \). His potential energy is all dependent on the height and mass, giving him an initial potential energy that "transfers" into the teeter-totter."
Teeter-totter Mechanism
A teeter-totter is a simple yet fascinating example of a lever in physics. It demonstrates how force and energy can be transferred from one end to the other.

In this scenario, when Bangles lands on one end, his gravitational potential energy gets converted into kinetic energy, which then transforms into potential energy as Juggles rises on the opposite end of the teeter-totter. This action can be broken down into two main processes:
  • Transfer of gravitational energy from Bangles to Juggles: The potential energy of Bangles is used to propel Juggles into the air.
  • Lever action: The teeter-totter acts as a fulcrum, where energy is redistributed causing Juggles to rise.
The teeter-totter beautifully illustrates principles of levers and energy conversion, maintaining balance through equal energy on both sides.
Conservation of Energy
The principle of conservation of energy is a foundational concept in physics. It states that energy cannot be created or destroyed, only transformed from one form to another.

In situations like our teeter-totter example, the initial potential energy Bangles possesses at the top is converted into various forms as he falls, eventually raising Juggles. The remarkable part is that despite these conversions, the total amount of energy remains constant.
  • Before the jump: All energy is potential in Bangles, with a value of \( 2,110.35 \text{ J} \).
  • After the jump: This energy raises Juggles, which we equate to find his mass.
In physics, understanding this principle allows us to solve for unknowns like Juggles' mass without additional energy sources鈥攕howcasing the balance and predictability in natural laws.

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Most popular questions from this chapter

"Rocket Man" has a propulsion unit strapped to his back. He starts from rest on the ground, fires the unit, and accelerates straight upward. At a height of \(16 \mathrm{m},\) his speed is \(5.0 \mathrm{m} / \mathrm{s} .\) His mass, including the propulsion unit, has the approximately constant value of 136 kg. Find the work done by the force generated by the propulsion unit.

Starting from rest, a 93-kg firefighter slides down a fire pole. The average frictional force exerted on him by the pole has a magnitude of \(810 \mathrm{N},\) and his speed at the bottom of the pole is \(3.4 \mathrm{m} / \mathrm{s}\). How far did he slide down the pole?

A 55.0-kg skateboarder starts out with a speed of 1.80 \(\mathrm{m} / \mathrm{s}\). He does \(+80.0 \mathrm{J}\) of work on himself by pushing with his feet against the ground. In addition, friction does -265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is \(6.00 \mathrm{m} / \mathrm{s} .\) (a) Calculate the change \(\left(\Delta \mathrm{PE}=\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{0}\right)\) in the gravitational potential energy. (b) How much has the vertical height of the skater changed, and is the skater above or below the starting point?

A small lead ball, attached to a 0.75-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is \(1.5 \mathrm{m}\) above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise?

A 75.0 -kg man is riding an escalator in a shopping mall. The escalator moves the man at a constant velocity from ground level to the floor above, a vertical height of \(4.60 \mathrm{m}\). What is the work done on the man by (a) the gravitational force and (b) the escalator?
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