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Chapter 6: Problem 30

Juggles and Bangles are clowns. Juggles stands on one end of a teetertotter 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

Calculate Bangles' Gravitational Potential Energy

To find the gravitational potential energy (GPE) that Bangles has before he jumps, use the formula for gravitational potential energy: \( PE = m \cdot g \cdot h \), where \( m = 86 \ kg \), \( g = 9.8 \ m/s^2 \) (the acceleration due to gravity), and \( h = 2.5 \ m \). Thus, \( PE_{Bangles} = 86 \times 9.8 \times 2.5 = 2107 \ J \).
02

Use Potential Energy Equation for Juggles

Since Juggles rises to a height of 3.3 meters, he has the same gravitational potential energy that Bangles had before jumping. Therefore, Bangles' potential energy is equal to Juggles' potential energy: \( PE_{Juggles} = m_{Juggles} \cdot g \cdot 3.3 \). Set \( PE_{Juggles} = 2107 \ J \) from Step 1.
03

Solve for Juggles' Mass

Set the equations from Bangles' and Juggles' potential energies equal: \( m_{Juggles} \cdot 9.8 \cdot 3.3 = 2107 \). Divide both sides by \( 9.8 \cdot 3.3 \). This gives \( m_{Juggles} = \frac{2107}{9.8 \times 3.3} \approx 65 \ kg \).

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Key Concepts

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

Teeter-totter physics
A teeter-totter, also known as a seesaw, is a simple lever that is used to demonstrate basic principles of physics. It consists of a plank balanced on a pivot point, or fulcrum, where two people can sit on opposite ends to create a balanced system. In physics, this setup exemplifies how forces and moments work together to create equilibrium鈥攐r lack thereof. In our clown scenario, Bangles jumping onto one end of the teeter-totter creates an imbalance of forces. The energy transferred from his jump is what allows Juggles to be lifted into the air. The teeter-totter essentially acts as a lever that emphasizes the transfer of mechanical energy. This example also provides a fun and practical way to explore concepts like equilibrium and rotational motion.
Conservation of Energy
The principle of Conservation of Energy is a cornerstone of physics, stating that energy cannot be created or destroyed in an isolated system. Instead, it can only change from one form to another. In our exercise with Juggles and Bangles, we deal specifically with the transformation between gravitational potential energy and kinetic energy.
  • Bangles has gravitational potential energy when he is on the platform, which converts into kinetic energy as he jumps down.
  • This kinetic energy is then transferred to Juggles through the teeter-totter, converting back into gravitational potential energy as Juggles rises into the air.
This constant transformation exemplifies the conservation law: the total energy in the system remains the same, merely switching forms.
Mass calculations in physics
Mass calculations are an important aspect of solving physics problems, particularly when dealing with gravitational systems. In our exercise, we calculated Juggles' mass by equating his potential energy to Bangles'. This process involved using the formula for gravitational potential energy: \[ PE = m \cdot g \cdot h \]This formula allowed us to find Juggles鈥� mass by knowing the gravitational potential energy provided by Bangles鈥� jump:
  • Set Bangles' potential energy equal to Juggles' potential energy.
  • Rearrange the formula to solve for mass: \[ m_{Juggles} = \frac{PE}{g \cdot h} \]
Mass calculation like this is crucial for understanding how different objects will interact under the force of gravity, and it highlights the importance of knowing an object's mass to predict its motion in various scenarios.

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Most popular questions from this chapter

A small lead ball, attached to a 1.5-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 \(0.75 \mathrm{m}\) 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 husband and wife take turns pulling their child in a wagon along a horizontal sidewalk. Each exerts a constant force and pulls the wagon through the same displacement. They do the same amount of work, but the husband's pulling force is directed \(58^{\circ}\) above the horizontal, and the wife's pulling force is directed \(38^{\circ}\) above the horizontal. The husband pulls with a force whose magnitude is \(67 \mathrm{N}\). What is the magnitude of the pulling force exerted by his wife?

A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising the helicopter. An \(810-\mathrm{kg}\) helicopter rises from rest to a speed of \(7.0 \mathrm{m} / \mathrm{s}\) in a time of \(3.5 \mathrm{s}\) During this time it climbs to a height of \(8.2 \mathrm{m}\). What is the average power generated by the lifting force?

Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the drive force is shut off, the snowmobile coasts to a halt. The snowmobile and its rider have a mass of \(136 \mathrm{kg}\). Under the influence of a drive force of \(205 \mathrm{N},\) it is moving at a constant velocity whose magnitude is \(5.50 \mathrm{m} / \mathrm{s}\) The drive force is then shut off. Find (a) the distance in which the snowmobile coasts to a halt and (b) the time required to do so.

A person is making homemade ice cream. She exerts a force of magnitude \(22 \mathrm{N}\) on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius \(0.28 \mathrm{m}\). The force is always applied parallel to the motion of the handle. If the handle is turned once every \(1.3 \mathrm{s},\) what is the average power being expended?
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