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Chapter 6: Problem 25

A 12 -pack of Omni-Cola (mass 4.30 \(\mathrm{kg}\) ) is initially at rest on a horizontal floor. It is then pushed in a straight line for 1.20 \(\mathrm{m}\) by a trained dog that exerts a horizontal force with magnitude 36.0 \(\mathrm{N}\) . Use the work-energy theorem to find the final speed of the 12-pack if (a) there is no friction between the 12 -pack and the floor, and (b) the coefficient of kinetic friction between the 12 -pack and the floor is \(0.30 .\)

Short Answer

Expert verified
(a) 4.46 m/s, (b) 3.62 m/s.

Step by step solution

01

Understanding the Problem

We need to find the final speed of a 12-pack pushed by a dog for 1.20 m. In part (a), there is no friction, while in part (b), there is kinetic friction with a coefficient of 0.30. The work-energy theorem states that the work done on an object is equal to its change in kinetic energy.
02

Calculating Work Done by the Dog (No Friction)

First, let's calculate the work done by the dog. Work done (W) is given by the formula \( W = F \times d \times \cos(\theta) \), where \( F = 36.0 \ \text{N} \), \( d = 1.20 \ \text{m} \), and \( \theta = 0^\circ \) because the force is horizontal. Thus, \( W = 36.0 \times 1.20 \times \cos(0) = 36.0 \times 1.20 = 43.2 \, \text{J} \).
03

Applying the Work-Energy Theorem (No Friction)

According to the work-energy theorem, the work done is equal to the change in kinetic energy: \( W = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \). Since the 12-pack starts from rest, \( u = 0 \). Therefore, \( 43.2 = \frac{1}{2} \times 4.30 \times v^2 \). Solving for \( v \), we get \( v = \sqrt{\frac{2 \times 43.2}{4.30}} \), which results in \( v \approx 4.46 \, \text{m/s} \).
04

Calculating Frictional Force (With Friction)

Now, for part (b), we consider friction. The frictional force (\( f_k \)) is calculated using \( f_k = \mu_k \times m \times g \), where \( \mu_k = 0.30 \), \( m = 4.30 \ \text{kg} \), and \( g = 9.8 \ \text{m/s}^2 \). Thus, \( f_k = 0.30 \times 4.30 \times 9.8 = 12.636 \, \text{N} \).
05

Calculating Net Work Done (With Friction)

The net work done is the work by the dog minus the work done against friction. Work against friction is \( f_k \times d = 12.636 \times 1.20 = 15.1632 \, \text{J} \). The net work done is \( 43.2 - 15.1632 = 28.0368 \, \text{J} \).
06

Applying the Work-Energy Theorem (With Friction)

Using the net work done and the work-energy theorem, \( 28.0368 = \frac{1}{2} \times 4.30 \times v^2 \). Solving for \( v \), we have \( v = \sqrt{\frac{2 \times 28.0368}{4.30}} \), which results in \( v \approx 3.62 \, \text{m/s} \).

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Key Concepts

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

Kinetic Friction
When objects slide against each other, they encounter resistance known as kinetic friction. This force opposes the motion of the object and acts parallel to the contact surface. In the case of moving a 12-pack of Omni-Cola across a floor with friction, we account for this resistance using the coefficient of kinetic friction \( \mu_k \). This dimensionless value represents how grippy two surfaces are when sliding past each other. The higher the \( \mu_k \), the more frictional force there is. Here, with \( \mu_k \) given as 0.30, friction significantly affects motion, slowing our 12-pack down.
Work Done
Work is an essential part of mechanics representing how much energy is transferred through force acting over a distance. Calculating work involves the formula \( W = F \times d \times \cos(\theta) \), where \( F \) is the force applied, \( d \) is the distance moved by the object, and \( \theta \) is the angle between the force and the direction of motion. For our exercise, since the force exerted by the dog is horizontal, \( \theta = 0^\circ \), making \( \cos(0^\circ) = 1 \). This simplifies our calculation to \( W = F \times d \). This quick computation assists in determining how much energy the dog transfers to the Omni-Cola packet, paving the way for determining kinetic energy changes.
Kinetic Energy
Kinetic energy is the energy of an object due to its motion, calculated using the equation \( KE = \frac{1}{2} m v^2 \), where \( m \) represents mass and \( v \) stands for velocity. As work is done on the Omni-Cola, its kinetic energy changes. Initially, it starts from rest, so its initial kinetic energy is zero. All work done goes into raising its kinetic energy. Thus, the calculation of kinetic energy helps us find out how fast the 12-pack moves after being pushed by the dog, given the conditions of friction or the lack thereof. Understanding kinetic energy variations explains the effects of net forces on the object's speed.
Net Work
Net work determines the total work done on an object after considering all forces, including friction. In this context, the net work is the initial work done by the dog minus the work associated with overcoming friction. Here's how it works:
- Calculate the work done by friction: \( f_k \times d \).
- Subtract this from the total work done by the dog to find the net work.
This net work practically determines the remaining energy converting into kinetic energy, influencing the speed of the object. It elucidates how various forces influence the system, shedding light on the power distribution and outcome of different friction scenarios.
Force Calculation
Calculating forces correctly is fundamental to understanding motion. The dog applies a force that drives the system, initiating the slide of the 12-pack. Use \( F = m \times a \), where \( F \) is the force, \( m \) is mass, and \( a \) is acceleration, to determine how much force is necessary to accelerate the Omni-Cola pack. In our situation, knowing the force exerted (36.0 N) is crucial as it accounts for the primary factor driving the system. In combination with frictional force, it dictates how effectively the 12-pack gains speed over the given distance. Mastery of force calculation reveals the required energy input and resistance encountered, mapping how different elements interplay in motion.

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Most popular questions from this chapter

A mass \(m\) slides down a smooth inclined plane from an initial vertical height \(h,\) making an angle \(\alpha\) with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height \(h\) . (b) Use the work-energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if it had been dropped from height \(h,\) independent of the angle \(\alpha\) of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy friction- less hill, starting from rest 15.0 m above the bottom.

The Grand Coulee Dam is 1270 \(\mathrm{m}\) long and 170 \(\mathrm{m}\) high. The electrical power output from generators at its base is approximately 2000 \(\mathrm{MW}\) . How many cubic meters of water must flow from the top of the dam per second to produce this amount of power if 92\(\%\) of the work done on the water by gravity is converted to electrical energy? (Each cubic meter of water has a mass of 1000 \(\mathrm{kg.})\)

Two blocks are connected by a very light string passing over a massless and frictionless pulley (Fig. E6.7). Traveling at constant speed, the 20.0 -N block moves 75.0 \(\mathrm{cm}\) to the right and the 12.0 -N block moves 75.0 \(\mathrm{cm}\) downward. During this process, how much work is done (a) on the \(12.0-\mathrm{N}\) block by (i) gravity and (ii) the tension in the string? (b) On the 20.0 -N block by (i) gravity, (ii) the tension in the string, (iii) friction, and (iv) the normal force? (c) Find the total work done on each block.

BIO Chin-Ups, While doing a chin-up, a man lifts his body 0.40 \(\mathrm{m} .\) (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 \(\mathrm{J}\) of work per kilogram of muscle mass. If the man can just barely do a \(0.40-\mathrm{m}\) chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical \(70-\mathrm{kg}\) man with 14\(\%\) body fat is about 43\(\%\) . (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 \(\mathrm{J}\) of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.

BIO Power of the Human Heart. The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 \(\mathrm{L}\) of blood. Assume that the work done by the heart is equal to the work required to lift this amount of blood a height equal to that of the average American woman \((1.63 \mathrm{m}) .\) The density (mass per unit volume) of blood is \(1.05 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) . (a) How much work does the heart do in a day? (b) What is the heart's power output in watts?
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