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A \(20.0-\mathrm{kg}\) rock is sliding on a rough, horizontal surface at 8.00 \(\mathrm{m} / \mathrm{s}\) and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is \(0.200 .\) What average power is produced by friction as the rock stops?

Short Answer

Expert verified
The average power produced by friction is approximately 156.86 W.

Step by step solution

01

Understand the problem

We have a rock with a given mass sliding on a horizontal surface, initially moving at a given speed, and it comes to a stop due to friction. We need to calculate the average power produced by the friction force.
02

Write down known values

Mass of the rock, \( m = 20.0 \, \text{kg} \), initial velocity, \( v_i = 8.00 \, \text{m/s} \), final velocity, \( v_f = 0 \, \text{m/s} \), and coefficient of kinetic friction, \( \mu_k = 0.200 \).
03

Calculate the force of friction

The force of friction is calculated using the formula: \( F_f = \mu_k \cdot m \cdot g \), where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. Thus, \( F_f = 0.200 \times 20.0 \times 9.81 = 39.24 \, \text{N} \).
04

Find the work done by friction

The work done by friction is equal to the change in kinetic energy of the rock. It can be calculated as \( W = \frac{1}{2} m v_i^2 \). Using \( m = 20.0 \, \text{kg} \) and \( v_i = 8.00 \, \text{m/s} \), we have \( W = \frac{1}{2} \times 20.0 \times 8.00^2 = 640.00 \, \text{J} \).
05

Calculate the distance over which the force acts

Using the work-energy principle, work done by friction \( W = F_f \times d \), where \( d \) is the distance. Solving for \( d \): \( 640.00 = 39.24 \times d \), \( d = \frac{640.00}{39.24} \approx 16.31 \, \text{m} \).
06

Determine the time taken to stop

Using the formula for constant acceleration, we have \( v_f = v_i + at \) to find \( t \), where \( a = \frac{F_f}{m} = \frac{39.24}{20.0} = -1.962 \, \text{m/s}^2 \). Solving for \( t \): \( 0 = 8 - 1.962t \), \( t = \frac{8}{1.962} \approx 4.08 \, \text{s} \).
07

Calculate the average power produced by friction

The average power can be calculated as \( P = \frac{W}{t} \), with \( W = 640.00 \, \text{J} \) and \( t = 4.08 \, \text{s} \). Therefore, \( P = \frac{640.00}{4.08} \approx 156.86 \, \text{W} \).

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Key Concepts

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

Kinetic Friction
Kinetic friction plays a pivotal role in the motion of objects when surfaces come into contact during movement. Specifically, kinetic friction is the force that opposes the relative motion or sliding of two surfaces. The strength of this force is determined by the coefficient of kinetic friction, which is a dimensionless number unique to the pair of materials in contact.

In this problem, the coefficient of kinetic friction given is 0.200, and the rock's mass is 20.0 kg, sliding over a horizontal surface. The kinetic friction force is computed using the formula: \[ F_f = \mu_k \times m \times g \]where \(\mu_k\) is the coefficient of kinetic friction, \(m\) is the mass, and \(g\) is the acceleration due to gravity (9.81 m/s²). This force ultimately causes the rock to come to a stop, representing an essential part of the overall physics problem.

Through consistent application of this formula, one observes how different surfaces and material properties can affect the motion, compressing the rock's energy into heat and other forms, which leads us into the application of the work-energy principle.
Work-Energy Principle
The work-energy principle is a cornerstone concept in physics, emphasizing the relationship between work done and the change in kinetic energy of an object. This principle states that the work done by the forces acting on an object results in a change in the object's kinetic energy. Here, we can use it to discern how friction slows and eventually halts a moving rock.

The work done by friction can be directly related to the change in kinetic energy through the formula:\[ W = \frac{1}{2}mv_i^2 \] where \(W\) is the work done (or the energy transferred by the force), \(m\) is the mass, and \(v_i\) is the initial velocity. Since the rock stops, the final kinetic energy is zero, thus simplifying calculations.

The computed work in this case is 640.00 J, showing the frictional force depleted the rock's kinetic energy over the course of 16.31 meters. This pivotal principle underlines the interdependence of force, distance, and energy transformation.
Power Calculation
In physics, power is not only related to how much work is done but also to how quickly that work is completed. This makes it an important measure in mechanical systems, especially when it involves forces like friction. The average power produced by friction can be calculated using:\[ P = \frac{W}{t} \]where \(P\) is the power, \(W\) is the work done, and \(t\) is the time taken.

For the rock in motion, having initially been in motion for 4.08 seconds and progressing until completely stopped, the average power output is found to be approximately 156.86 Watts. This value reflects how effectively the frictional forces convert the kinetic energy of the rock into other energy forms such as heat.

Understanding power in this framework helps connect the dots between energy, time, and force, allowing further examination into the efficiency and role of forces in dynamic systems.

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Most popular questions from this chapter

The spring of a spring gun has force constant \(k=400 \mathrm{N} / \mathrm{m}\) and negligible mass. The spring is compressed \(6.00 \mathrm{cm},\) and a ball with mass 0.0300 \(\mathrm{kg}\) is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 \(\mathrm{cm}\) long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. (a) Calculate the speed with which the ball leaves the barrel if you can ignore friction. (b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 \(\mathrm{N}\) acts on the ball as it moves along the barrel. (c) For the situation in part (b), at what position along the barrel does the ball have the greatest speed, and what is that speed? (In this case, the maximum speed does not occur at the end of the barrel)

Rotating Bar. A thin, uniform \(12.0-\mathrm{kg}\) bar that is 2.00 \(\mathrm{m}\) long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar? (Hint: Different points in the bar have different speeds. Break the bar up into infinitesimal segments of mass \(d m\) and integrate to add up the kinetic energy of all these segments.)

An ingenious bricklayer builds a device for shooting bricks up to the top of the wall where he is working. He places a brick on a vertical compressed spring with force constant \(k=450 \mathrm{N} / \mathrm{m}\) and negligible mass. When the spring is released, the brick is propelled upward. If the brick has mass 1.80 \(\mathrm{kg}\) and is to reach a maximum height of 3.6 \(\mathrm{m}\) above its initial position on the compressed spring, what distance must the bricklayer compress the spring initially? (The brick loses contact with the spring when the spring returns to its uncompressed length. Why?)

A soccer ball with mass 0.420 \(\mathrm{kg}\) is initially moving with speed 2.00 \(\mathrm{m} / \mathrm{s}\) . A soccer player kicks the ball, exerting a constant force of magnitude 40.0 \(\mathrm{N}\) in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 \(\mathrm{m} / \mathrm{s} ?\)

Automotive Power II. (a) If 8.00 hp are required to drive a \(1800-\mathrm{kg}\) automobile at 60.0 \(\mathrm{km} / \mathrm{h}\) on a level road, what is the total retarding force due to friction, air resistance, and so on? (b) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) up a 10.0\(\%\) grade (a hill rising 10.0 \(\mathrm{m}\) vertically in 100.0 \(\mathrm{m}\) horizontally)? (c) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) down a 1.00\(\%\) grade? (d) Down what percent grade would the car coast at 60.0 \(\mathrm{km} / \mathrm{h} ?\)

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