/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 127 In a circus act, a \(60 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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In a circus act, a \(60 \mathrm{~kg}\) clown is shot from a cannon with an initial velocity of \(16 \mathrm{~m} / \mathrm{s}\) at some unknown angle above the horizontal. A short time later the clown lands in a net that is \(3.9 \mathrm{~m}\) vertically above the clown's initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?

Short Answer

Expert verified
The kinetic energy of the clown as he lands is 5386.8 J.

Step by step solution

01

Understand the Problem

We need to find the kinetic energy of the clown when he lands in the net, which is 3.9 meters above his initial position. We're given the initial velocity of the clown as he is projected from the cannon, and we're asked to disregard air drag.
02

Apply the Conservation of Energy Principle

In the absence of air drag, mechanical energy is conserved. This means the sum of the initial kinetic energy and gravitational potential energy of the clown should equal the sum of the final kinetic energy and potential energy. We can set up the equation: \[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \] The initial potential energy \( PE_{initial} \) is zero because the clown starts from the ground. We are tasked to find \( KE_{final} \).
03

Calculate Initial Kinetic Energy

The initial kinetic energy \( KE_{initial} \) is given by the formula \[ KE_{initial} = \frac{1}{2}mv^2 \]where \( m = 60 \, \text{kg} \) and \( v = 16 \, \text{m/s} \). Substituting these values, we find \[ KE_{initial} = \frac{1}{2} \times 60 \times 16^2 = 7680 \, \text{J} \].
04

Calculate Final Potential Energy

The final gravitational potential energy \( PE_{final} \) is given by \[ PE_{final} = mgh \] where \( g = 9.8 \, \text{m/s}^2 \) and \( h = 3.9 \, \text{m} \). Thus,\[ PE_{final} = 60 \times 9.8 \times 3.9 = 2293.2 \, \text{J} \].
05

Solve for Final Kinetic Energy

Substitute the known values into the conservation of energy equation \[ 7680 + 0 = KE_{final} + 2293.2 \].Solving for \( KE_{final} \) gives:\[ KE_{final} = 7680 - 2293.2 = 5386.8 \, \text{J} \].

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Key Concepts

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

Conservation of Energy
The Conservation of Energy is a fundamental principle in physics that tells us energy can neither be created nor destroyed. However, energy can change forms. In the problem of the clown being shot from a cannon, different types of energy are involved, such as kinetic and potential energy. If we imagine the clown's trajectory:
- Initially, he has a large amount of kinetic energy due to his speed.
- As he rises to the net 3.9 meters above, some kinetic energy is transformed into potential energy.
- Upon reaching the highest point, the kinetic energy is at its minimum because his speed decreases, while potential energy is at its maximum.

The total mechanical energy remains constant in this system because air drag is ignored. Thus, we use the conservation of energy to determine the clown’s kinetic energy when he lands in the net.
The equation we use is:\[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \]Here, - \( KE \) stands for kinetic energy, - \( PE \) stands for potential energy.
Start with what energy the clown had, and convert it as needed throughout the act.
See how this principle helps unravel the problem and gives us the result.
Mechanical Energy
Mechanical Energy combines both kinetic and potential energy within a system. In this exercise, it represents the total energy of the clown throughout his flight. Mechanical energy can change forms, but the total amount stays the same (assuming no energy loss to friction or air resistance).
In our circus act case:
  • **Kinetic Energy** is the energy of motion. The clown initially has this kind of energy due to his launch velocity.
  • **Potential Energy** is stored energy due to position. It increases as the clown rises in height after being shot from the cannon.

When solving problems that involve mechanical energy, it is essential to account for both kinetic and potential energies at both the start and end points of the motion. The conservation of mechanical energy allows us to track how one form of energy changes into another while keeping the total quantity constant.
Potential Energy
Potential Energy in physics refers to the energy that is stored in an object because of its position relative to other objects. In our exercise, the clown's potential energy is all about his height above the ground. The higher he goes, the more potential energy he has.
  • It depends on three factors: the clown's mass (which is constant), the height reached, and the gravitational acceleration (approximately \(9.8 \, \text{m/s}^2\) on Earth).
  • The formula used is \( \text{PE} = mgh \), where \( m \) is mass, \( g \) is gravity, and \( h \) is height.
The potential energy is zero at the initial starting point. But when he reaches 3.9 meters, it is no longer zero, and we quantify it as \(2293.2 \, \text{J}\). This value plays a crucial role in finding the final kinetic energy by using the conservation of energy principle.
Physics Education
Physics education aims to deepen understanding of both abstract concepts and real-world phenomena. Applying these ideas to varied problems, like the clown shot from a cannon, sharpens critical thinking and problem-solving skills. For students:
  • Experiments and real-life problem examples make learning engaging.
  • The use of clear calculations and consistent equations demonstrates how physics principles manifest in everyday actions.
When teaching concepts such as kinetic and potential energy, real-world applications, like circus acts or roller coasters, help connect theory with practice.

Effective physics education often involves using approachable language, visuals and examples that bridge the gap between abstract principles and tangible experiences. Building a solid grasp on energy conservation concepts in fun settings promotes a deeper appreciation for the subject's beauty and complexity.

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Most popular questions from this chapter

The summit of Mount Everest is \(8850 \mathrm{~m}\) above sea level. (a) How much energy would a \(90 \mathrm{~kg}\) climber expend against the gravitational force on him in climbing to the summit from sea level? (b) How many candy bars, at \(1.25 \mathrm{MJ}\) per bar, would supply an energy equivalent to this? Your answer should suggest that work done against the gravitational force is a very small part of the energy expended in climbing a mountain.

A single conservative force \(F(x)\) acts on a \(1.0 \mathrm{~kg}\) particle that moves along an \(x\) axis. The potential energy \(U(x)\) associated with \(F(x)\) is given by $$ U(x)=-4 x e^{-x / 4} \mathrm{~J} $$ where \(x\) is in meters. At \(x=5.0 \mathrm{~m}\) the particle has a kinetic energy of \(2.0 \mathrm{~J} .\) (a) What is the mechanical energy of the system? (b) Make a plot of \(U(x)\) as a function of \(x\) for \(0 \leq x \leq 10 \mathrm{~m}\), and on the same graph draw the line that represents the mechanical energy of the system. Use part (b) to determine (c) the least value of \(x\) the particle can reach and (d) the greatest value of \(x\) the particle can reach. Use part (b) to determine (e) the maximum kinetic energy of the particle and (f) the value of \(x\) at which it occurs. (g) Determine an expression in newtons and meters for \(F(x)\) as a function of \(x\). (h) For what (finite) value of \(x\) does \(F(x)=0 ?\)

What is the spring constant of a spring that stores 25 J of elastic potential energy when compressed by 7.5 cm?

The luxury liner Queen Elizabeth 2 has a diesel-electric power plant with a maximum power of \(92 \mathrm{MW}\) at a cruising speed of \(32.5\) knots. What forward force is exerted on the ship at this speed? \((1 \mathrm{knot}=1.852 \mathrm{~km} / \mathrm{h} .)\)

A swimmer moves through the water at an average speed of \(0.22 \mathrm{~m} / \mathrm{s}\). The average drag force is \(110 \mathrm{~N}\). What average power is required of the swimmer?

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