/*! 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 64 In moving a 35.0 -kg desk from o... [FREE SOLUTION] | 91Ó°ÊÓ

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In moving a 35.0 -kg desk from one side of a classroom to the other, a professor finds that a horizontal force of \(275 \mathrm{~N}\) is necessary to set the desk in motion, and a force of \(195 \mathrm{~N}\) is necessary to keep it in motion at a constant speed. What are the coefficients of (a) static and (b) kinetic friction between the desk and the floor?

Short Answer

Expert verified
Static friction coefficient: 0.802; Kinetic friction coefficient: 0.569.

Step by step solution

01

Understand the Problem Statement

We are given two pieces of information: the force required to set the desk in motion (\( F_s = 275 \text{ N} \)) and the force required to keep it moving at a constant speed (\( F_k = 195 \text{ N} \)). The mass of the desk is given as \( m = 35.0 \text{ kg} \). We need to find the coefficients of static and kinetic friction.
02

Calculate the Normal Force

The normal force (\( N \)) is the force exerted by a surface to support the weight of an object resting on it. Because the desk is on a horizontal surface and there is no vertical movement, the normal force equals the gravitational force on the desk, calculated as:\[ N = m \, g \]Where \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity.Plugging in the values: \[ N = 35.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 343 \text{ N} \]
03

Calculate the Coefficient of Static Friction

The coefficient of static friction (\( \mu_s \)) is calculated using the formula:\[ F_s = \mu_s \, N \]Rearranging for \( \mu_s \): \[\mu_s = \frac{F_s}{N} = \frac{275 \, \text{N}}{343 \, \text{N}} \approx 0.802 \]
04

Calculate the Coefficient of Kinetic Friction

The coefficient of kinetic friction (\( \mu_k \)) is calculated using the formula:\[ F_k = \mu_k \, N \]Rearranging for \( \mu_k \):\[\mu_k = \frac{F_k}{N} = \frac{195 \, \text{N}}{343 \, \text{N}} \approx 0.569 \]
05

Conclusion

The coefficients of static and kinetic friction are thus approximately 0.802 and 0.569, respectively.

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Key Concepts

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

Static Friction
Static friction is the force that resists the initial movement of an object that is at rest. It acts parallel to the surfaces in contact. You can think of static friction as the force you must overcome to get something moving. In our problem, the force needed to start moving the desk was 275 N. This force identifies the maximum force of static friction.
  • This force happens because surfaces are never perfectly smooth, leading to tiny rough patches that "lock" together.
  • Static friction prevents objects from moving until a threshold force is applied.
To find the coefficient of static friction (\(\mu_s\)), we use the formula: \[ F_s = \mu_s \cdot N \]Here, \(F_s\) is the force of static friction, and \(N\) is the normal force. In our task, calculating \(\mu_s\) gives us approximately 0.802. This coefficient is a ratio, indicating that the resisting frictional force is 80.2% of the normal force.
Kinetic Friction
While static friction involves getting an object moving, kinetic friction happens once the object is already in motion. In this situation, the kinetic friction is slightly less than static friction. It resists movement rather than stopping it from starting. In our example, a 195 N force was needed to keep the desk moving at a constant speed.
  • Kinetic friction is often lower than static friction because there is less "locking" between surfaces once they are sliding past one another.
  • This friction always acts opposite to the direction of motion.
To find the coefficient of kinetic friction (\(\mu_k\)), you apply:\[ F_k = \mu_k \cdot N \]By solving this equation, we obtained \(\mu_k \approx 0.569\). This means that kinetic friction is about 56.9% of the normal force.Understanding kinetic friction helps predict how much force is needed to maintain movement of an object.
Normal Force
The normal force (\(N\)) is crucial to solving friction problems. It is the force exerted by a surface to support the weight of an object resting on it. It's always perpendicular to the contact surface. In physics, it balances the gravitational force acting on an object on a horizontal surface, which explains why objects don’t fall through.
  • The weight of the desk creates a downward force due to gravity.
  • The normal force counteracts this force to prevent the desk from accelerating vertically.
In our problem, using the formula \(N = m \cdot g\) where \(m = 35.0 \,\text{kg}\) (mass of the desk) and \(g = 9.8 \, \text{m/s}^2\) (gravitational acceleration), we calculated \(N = 343 \, \text{N}\). This value is essential for determining both static and kinetic forces.
Physics Problem Solving
Solving physics problems is like a puzzle. It requires understanding concepts and applying them appropriately. For friction problems, understanding the types of friction and calculating the normal force is key.
  • Begin by identifying forces involved, like in our task where static and kinetic friction were primary forces of focus.
  • Recognize that balancing the equations of motion is fundamental. It helps to determine unknowns using known quantities.

In our specific example, the problem required calculating coefficients of friction. This involved translating word problems into mathematical equations, reflecting the physics laws in numbers. You must adapt your approach according to the context, knowing when and how to use particular formulas.
  • Normal calculations occur frequently and help isolate other forces acting on a body.
  • You'll use logic and algebra to rearrange formulas and find desired results.
Ultimately, problem-solving builds a deeper insight into how physical principles interplay with real-world scenarios.

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Most popular questions from this chapter

A block is projected with a speed of \(3.0 \mathrm{~m} / \mathrm{s}\) on a horizontal surface. If the coefficient of kinetic friction between the block and the surface is 0.60 , how far does the block slide before coming to rest?

In loading a fish delivery truck, a person pushes a block of ice up a \(20^{\circ}\) incline at constant speed. The push is \(150 \mathrm{~N}\) in magnitude and parallel to the incline. The block has a mass of \(35.0 \mathrm{~kg}\). (a) Is the incline frictionless? (b) If not, what is the force of kinetic friction on the block of ice?

A block of mass \(25.0 \mathrm{~kg}\) slides down a frictionless surface inclined at \(30^{\circ} .\) To ensure that the block does not accelerate, what is the smallest force that you must exert on it and what is its direction?

When a horizontal force of \(300 \mathrm{~N}\) is applied to a 75.0 \(\mathrm{kg}\) box, the box slides on a level floor, opposed by a force of kinetic friction of \(120 \mathrm{~N}\). What is the magnitude of the acceleration of the box?

IE .?? Three horizontal forces (the only horizontal ones) act on a box sitting on a floor. One (call it \(F_{1}\) ) acts due east and has a magnitude of \(150 \mathrm{lb}\). A second force (call it \(F_{2}\) ) has an easterly component of \(30.0 \mathrm{lb}\) and a southerly component of \(40.0 \mathrm{lb}\). The box remains at rest. (Neglect friction.) (a) Sketch the two known forces on the box. In which quadrant is the unknown third force: (1) the first quadrant; (2) the second quadrant; (3) the third quadrant; or (4) the fourth quadrant? (b) Find the unknown third force in newtons and compare your answer to the sketched estimate.

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