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Wheels. You find that it takes a horizontal force of 160 \(\mathrm{N}\) to slide a box along the surface of a level fioor at constant speed. The coefficient of static friction is \(0.52,\) and the coefficient of kinetic friction is \(0.47 .\) If you place the box on a dolly of mass 5.3 \(\mathrm{kg}\) and with coefficient of rolling friction \(0.018,\) what horizontal acceleration would that \(160-\mathrm{N}\) force provide?

Short Answer

Expert verified
The horizontal acceleration is approximately 30.02 m/s².

Step by step solution

01

Understand the forces involved

To find the acceleration, we need to consider all the forces acting on the system when the box is placed on the dolly. The force needed to overcome rolling friction and the effective force available for acceleration needs to be calculated.
02

Calculate the force of rolling friction

The rolling friction force, \( F_{r} \), can be calculated using the formula: \( F_{r} = \mu_{r} \times m \times g \), where \( \mu_{r} = 0.018 \) (coefficient of rolling friction), \( m = 5.3 \ kg \) (mass of the dolly), and \( g = 9.8 \ m/s^2 \) (acceleration due to gravity). This gives \( F_{r} = 0.018 \times 5.3 \times 9.8 \approx 0.93 \ N \).
03

Calculate the net force for acceleration

The net force available for acceleration is the original force minus the rolling friction force: \( F_{net} = F - F_{r} = 160 \ N - 0.93 \ N = 159.07 \ N \).
04

Calculate the acceleration of the system

Use Newton's second law to calculate the acceleration, \( a \). The formula is \( F_{net} = m_{total} \times a \), where \( m_{total} \) is the total mass of the system. Let's assume the mass of the box is very small compared to the dolly, so \( m_{total} = 5.3 \ kg \). Therefore, \( 159.07 \ N = 5.3 \ kg \times a \). Solving for \( a \) gives \( a = \frac{159.07}{5.3} \approx 30.02 \ m/s^2 \).

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Key Concepts

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

Rolling Friction
Rolling friction is a resistance encountered by any object that rolls over a surface. It is usually much smaller than other types, like static or kinetic friction, because rolling typically involves less surface contact. When a box is placed on a dolly, rolling friction becomes the primary force to overcome rather than sliding friction you might encounter if just dragging the box.

The rolling friction force can be calculated with a straightforward formula: \[ F_{r} = \mu_{r} \times m \times g \]where \( \mu_{r} \) is the coefficient of rolling friction, \( m \) is the mass of the object (like the dolly in this exercise), and \( g \) is the acceleration due to gravity. While the coefficient of rolling friction is typically a small number, it plays a significant role when calculating how much force you need for movement. This concept illustrates why wheels, often subject to rolling friction, are used in transportation over sliding to reduce energy needed to move objects.
Newton's Second Law
Newton's Second Law is one of the foundational principles in physics. It states that the acceleration of an object is directly proportional to the net force acting upon it, and inversely proportional to its mass. This can be represented with the formula:\[ F_{net} = m_{total} \times a \]where \( F_{net} \) is the net force applied on the object, \( m_{total} \) is the total mass, and \( a \) is the acceleration.

In the context of the exercise, after deducing the rolling friction, the net force available for acceleration is calculated by subtracting the rolling friction from the total applied force. The net force is then used in conjunction with the total mass to find the acceleration of the system using Newton's Second Law. This demonstrates how Newton's law links force and motion and provides a way to predict how objects will move when forces are applied to them.
Kinetic Friction
Kinetic friction, also known as dynamic friction, comes into play when two surfaces are moving relative to each other. Unlike static friction, which prevents motion, kinetic friction acts against moving objects and is often slightly less than static friction. The formula to calculate it is similar to static friction:\[ F_k = \mu_k \times N \]where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force, typically equal to the weight of the object if on a horizontal surface.

In the original exercise scenario, prior to using the dolly, kinetic friction dictates how much force you need to keep the box sliding at constant speed. Compared to rolling friction, kinetic friction usually requires more force to overcome because sliding surfaces interact more closely than rolling surfaces. By reducing kinetic friction through rolling with a dolly, the problem becomes easier to solve, illustrating why wheels reduce energy expenditure in motion.

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Most popular questions from this chapter

A 50.0-kg stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. (a) If the plane's speed at the lowest point of the circle is 95.0 \(\mathrm{m} / \mathrm{s}\) , what is the minimum radius of the circle for the acceleration at this point not to exceed 4.00 \(\mathrm{g} ?\) (b) What is the apparent weight of the pilot at the lowest point of the pullout?

Another Rope with Mass. A block with mass \(M\) is attached to the lower end of a vertical, uniform rope with mass \(m\) and length \(L\) . A constant upward force \(\vec{F}\) is applied to the top of the rope, causing the rope and block to accelerate upward. Find the tension in the rope at a distance \(x\) from the top end of the rope, where \(x\) can have any value from 0 to \(L\) .

Two objects with masses 5.00 \(\mathrm{kg}\) and 2.00 \(\mathrm{kg}\) hang 0.600 \(\mathrm{m}\) above the floor from the ends of a cord 6.00 \(\mathrm{m}\) long passing over a frictionless pulley. Both objects start from rest. Find the maximum height reached by the \(2.00-\mathrm{kg}\) object.

Free-Body Diagrams. The first two steps in the solution of Newton's second-law problems are to select an object for analysis and then to draw frec-body diagrams for that object. Draw free-body diagrams for the following situations: (a) a mass \(M\) sliding down a frictionless inclined plane of angle \(\alpha,\) and (b) a mass \(M\) sliding up a frictionless inclined plane of angle \(\alpha ;\) (c) a mass \(M\) sliding up an inclined plane of angle \(\alpha\) with kinetic friction present.

Stopping Distance. (a) If the coefficient of kinctic friction between tires and dry pavement is 0.80 , what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 28.7 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} ) ?\) (b) On wet pavement the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

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