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Chapter 3: Problem 19

A moving object whose initial KE is \(10 \mathrm{J}\) is subject to a frictional force of \(2 \mathrm{N}\) that acts in the opposite direction. How far will the object move before coming to a stop?

Short Answer

Expert verified
The object will move 5 meters before coming to a stop.

Step by step solution

01

Identify Given Information

The initial kinetic energy (KE) of the moving object is given as 10 J. The frictional force acting on the object is 2 N.
02

Understand the Concept

The object moves and is slowed down by the frictional force until it stops. The work done by the frictional force will equal the initial kinetic energy of the object. The work-energy principle states that the work done on the object will be equal to the change in its kinetic energy.
03

Calculate the Work Done by Friction

The work done (W) by the frictional force can be calculated using the formula: \[ W = F \times d \] where \( F \) is the frictional force and \( d \) is the distance over which the force acts.
04

Set Up the Equation

Since the work done by the frictional force is equal to the initial kinetic energy, set up the equation: \[ F \times d = 10 \text{ J} \]
05

Solve for Distance

Rearrange the equation to solve for \( d \): \[ d = \frac{10 \text{ J}}{2 \text{ N}} \] \[ d = 5 \text{ m} \]

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. The formula for kinetic energy is given by: \[ KE = \frac{1}{2} mv^2 \]where 'm' is the mass of the object and 'v' is its velocity. In our exercise, the initial kinetic energy of the moving object is 10 Joules (J). This means that when the object is in motion, it has a certain amount of energy due to its motion.

Understanding kinetic energy helps in knowing how forces like friction will act to stop the object. The initial kinetic energy is converted into other energy forms due to forces, mainly friction in this case.
Frictional Force
Frictional force is a force that opposes the motion of an object. It acts in the opposite direction to the movement. This force results in the object slowing down. Friction is necessary for practical purposes like stopping cars or preventing slipping, but it also results in energy loss in the form of heat.

In our exercise, the frictional force acting on the object is 2 Newtons (N). This force will work against the kinetic energy and eventually bring the object to a stop. Understanding friction is essential for calculating how far the object will travel before stopping.
Work-Energy Principle
The work-energy principle is a vital concept in physics. It states that the work done on an object is equal to the change in its kinetic energy. The formula for work (W) done by a force (F) acting over a distance (d) is:

\[ W = F \times d \]Here, the work done by the frictional force will be equal to the initial kinetic energy of the object when it comes to a stop. So in the exercise, the force (2 N) multiplied by the distance (d) the object travels should equal the initial kinetic energy (10 J). This principle helps us set up the equation to find the distance.
Distance Calculation
To find the distance the object will travel before it stops, we use the work-energy principle. We've established that the work done by the frictional force is equal to the initial kinetic energy. Using the equation:

\[ F \times d = 10 \text{ J} \]
To find 'd' (distance), rearrange the equation:

\[ d = \frac{10 \text{ J}}{2 \text{ N}} \]
Solving this, we get:

\[ d = 5 \text{ m} \]
Therefore, the object will travel 5 meters before coming to a stop. Understanding this calculation is crucial for solving similar problems involving kinetic energy, friction, and the work-energy principle.

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Most popular questions from this chapter

A \(40-\mathrm{kg}\) skater moving at \(4 \mathrm{m} / \mathrm{s}\) overtakes a \(60-\mathrm{kg}\) skater moving at \(2 \mathrm{m} / \mathrm{s}\) in the same direction and collides with her. The two skaters stick together (a) What is their final speed? (b) How much kinetic energy is lost?

A \(70-\mathrm{kg}\) man and a 50 -kg woman are in a 60 -kg boat when its motor fails. The man dives into the water with a horizontal speed of \(3 \mathrm{m} / \mathrm{s}\) in order to swim ashore. If he changes his mind, can he swim back to the boat if his swimming speed is \(1 \mathrm{m} / \mathrm{s}\) ? If not can the woman change the boat's motion enough by diving off it at \(3 \mathrm{m} / \mathrm{s}\) in the opposite direction? Could she then return to the boat herself if her swimming speed is also \(1 \mathrm{m} / \mathrm{s}\) ?

A ball is dropped from a height of \(1 \mathrm{m}\) and loses 10 percent of its kinetic energy when it bounces on the ground. To what height does it then rise?

A 700 -kg horse whose power output is 1.0 hp is pulling a sled over the snow at \(3.5 \mathrm{m} / \mathrm{s}\). Find the force the horse exerts on the sled.

As we will learn in Chap. \(6,\) electric charges of the same kind (both positive or both negative) repel each other, whereas charges of opposite sign (one positive and the other negative) attract each other (a) What happens to the PE of a positive charge when it is brought near another positive charge? (b) When it is brought near a negative charge?
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