/*! 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 27 Two workers are sliding a \(300 ... [FREE SOLUTION] | 91Ó°ÊÓ

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Two workers are sliding a \(300 \mathrm{kg}\) crate across the floor. One worker pushes forward on the crate with a force of \(380 \mathrm{N}\) while the other pulls in the same direction with a force of \(350 \mathrm{N}\) using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?

Short Answer

Expert verified
The crate's coefficient of kinetic friction on the floor is 0.248.

Step by step solution

01

Calculate the Total Force Applied by the Workers

Since both workers are applying forces in the same direction (i.e., they are parallel and co-directional), adding their forces will give the total force applied on the crate. Therefore, the total force \(F\) is equal to \(380N + 350N = 730N\)
02

Calculate the Normal Force

Normal force \(N\) is the force exerted by a surface that supports the weight of an object on it. It acts perpendicular to the surface. In this case, since the crate is not moving vertically, the normal force is equal to the weight of the crate which is the mass of the crate \(m\) times the gravitational acceleration \(g = 9.8 m/s^2\). Hence, \(N = m \times g = 300kg \times 9.8m/s^2 = 2940N\)
03

Determine the Coefficient of Kinetic Friction

Since the crate is moving at a constant speed, it means that the total horizontal force applied on the crate by the workers is equal to the force of kinetic friction \(f_{k}\). Therefore, using the equation \(f_{k} = \mu_{k} \times N\) where \(f_{k}\) is the force of kinetic friction and \(N\) is the normal force, we can solve for the coefficient of kinetic friction \(\mu_{k}\). \(\mu_{k} = f_{k} / N = 730N / 2940N = 0.248\)

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Key Concepts

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

Newton's Laws of Motion
Understanding Newton's Laws of Motion is fundamental to solving problems related to forces and motion. Newton's First Law, sometimes called the Law of Inertia, states that an object will remain at rest or move with a constant velocity unless acted upon by a net external force. In the problem of the sliding crate, since it moves at a constant speed, the net force acting on it is zero. This means that the forces applied by the workers are perfectly balanced by the kinetic friction force. Newton's Second Law ties force, mass, and acceleration together, expressed by the equation \( F = ma \). In our scenario, since the crate moves at constant speed, the acceleration \( a \) is zero, implying the net force is zero. Lastly, Newton's Third Law tells us that for every action, there is an equal and opposite reaction, which helps us understand how the floor pushes back with a normal force.
Normal Force
The normal force is a key concept in understanding how objects interact with surfaces. It is the perpendicular force exerted by a surface to support the weight of an object resting on it. In this exercise, the crate is resting on a horizontal floor, meaning there are no vertical movements. Because of this, the normal force \( N \) is equal to the gravitational force acting on the crate. This can be calculated by multiplying the mass of the crate \( m \) by the gravitational acceleration \( g \), resulting in \( N = m \times g = 300 \text{ kg} \times 9.8 \text{ m/s}^2 = 2940 \text{ N} \). The normal force plays a crucial role in kinetic friction calculations because it influences the magnitude of the frictional force.
Kinetic Friction
Kinetic friction occurs when two surfaces slide past each other. It resists the motion, and its magnitude can be calculated once you know the normal force and the coefficient of kinetic friction. The force of kinetic friction \( f_k \) is given by \( f_k = \mu_k \times N \), where \( \mu_k \) is the coefficient of kinetic friction. In the problem given, both workers apply a combined force to keep the crate moving at constant speed, balancing the kinetic friction force. To find \( \mu_k \), you divide the total applied force by the normal force, resulting in \( \mu_k = \frac{730 \text{ N}}{2940 \text{ N}} = 0.248 \). This coefficient is a measure of how easily the crate slides over the floor.
Constant Speed
Moving at constant speed indicates a balance in forces, meaning the net force on an object is zero. In the practice problem, this balance is achieved between the applied force by the workers and the kinetic frictional force opposing it. Since the speed is constant, the total horizontal force applied equals the kinetic friction force. This concept is vital for understanding the problem because it allows us to equate the total force exerted by the workers to the friction force. By doing so, we can accurately determine the coefficient of kinetic friction, as all forces have reached equilibrium. Recognizing when forces are in balance helps simplify these types of physics problems significantly.

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Most popular questions from this chapter

Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first block is pulled with a force of \(100 \mathrm{N}\). a. What is the tension in the string connecting block 100 to block 99? b. What is the tension in the string connecting block 50 to block \(51 ?\)

A simple model shows how drawing a bow across a violin string causes the string to vibrate. As the bow moves across the string, static friction between the bow and the string pulls the string along with the bow. At some point, the tension pulling the string back exceeds the maximum static friction force and the string snaps back. This process repeats cyclically, causing the string's vibration. Assume the tension in a 0.33 -m-long violin string is \(50 \mathrm{N}\), and the coefficient of static friction between the bow and the string is \(\mu_{\mathrm{s}}=0.80 .\) If the normal force of the bow on the string is \(0.75 \mathrm{N},\) how far can the string be pulled before it slips if the string is bowed at its center?

A fisherman has caught a very large, \(5.0 \mathrm{kg}\) fish from a dock that is \(2.0 \mathrm{m}\) above the water. He is using lightweight fishing line that will break under a tension of \(54 \mathrm{N}\) or more. He is eager to get the fish to the dock in the shortest possible time. If the fish is at rest at the water's surface, what's the least amount of time in which the fisherman can raise the fish to the dock without losing it?

Riders on the Tower of Doom, an amusement park ride, experience \(2.0 \mathrm{s}\) of free fall, after which they are slowed to a stop in 0.50 s. What is a 65 kg rider's apparent weight as the ride is coming to rest? By what factor does this exceed her actual weight?

The acceleration of the spacecraft in which the Apollo astronauts took off from the moon was \(3.4 \mathrm{m} / \mathrm{s}^{2} .\) On the moon, \(g=1.6 \mathrm{m} / \mathrm{s}^{2} .\) What was the apparent weight of a \(75 \mathrm{kg}\) astronaut during takeoff?

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