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SSM ILW A person pushes horizontally with a force of \(220 \mathrm{~N}\) on a \(55 \mathrm{~kg}\) crate to move it across a level floor. The coefficient of kinetic friction is \(0.35 .\) What is the magnitude of (a) the frictional force and (b) the crate's acceleration?

Short Answer

Expert verified
(a) 188.65 N frictional force; (b) 0.57 m/s² acceleration.

Step by step solution

01

Determine the Normal Force

Since the crate is being pushed horizontally, the normal force \( N \) is equal to the weight of the crate. The weight \( W \) can be calculated as \( W = m \cdot g \), where \( m = 55 \; \text{kg} \) is the mass of the crate and \( g = 9.8 \; \text{m/s}^2 \) is the acceleration due to gravity. Thus, \( N = 55 \; \text{kg} \times 9.8 \; \text{m/s}^2 = 539 \; \text{N} \).
02

Calculate the Frictional Force

The frictional force \( f_k \) opposing the motion can be calculated using the formula \( f_k = \mu_k \cdot N \), where \( \mu_k = 0.35 \) is the coefficient of kinetic friction. Substitute the normal force calculated in Step 1: \( f_k = 0.35 \times 539 \; \text{N} = 188.65 \; \text{N} \).
03

Apply Newton's Second Law to Find Acceleration

To find the acceleration \( a \) of the crate, use the net force acting on it. The net force \( F_{net} \) is the applied force minus the frictional force: \( F_{net} = 220 \; \text{N} - 188.65 \; \text{N} = 31.35 \; \text{N} \). According to Newton's Second Law, \( F_{net} = m \cdot a \), solve for \( a \): \( a = \frac{F_{net}}{m} = \frac{31.35 \; \text{N}}{55 \; \text{kg}} \approx 0.57 \; \text{m/s}^2 \).

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle in physics that explains how the velocity of an object changes when it is subjected to external forces. The law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, represented by the equation:
  • \( F = m \cdot a \)
Here, \( F \) is the net force acting on the object, \( m \) is the mass, and \( a \) is the acceleration. This relationship implies that an increase in force will result in a greater acceleration if mass is constant. Similarly, for a given force, a larger mass will result in smaller acceleration.
In our exercise, a force of \(220 \text{ N}\) is applied to the crate. The net force, accounting for friction, is calculated and used to determine the acceleration of the crate using Newton's Second Law. It provides a clear understanding of how forces and mass interact to influence motion.
Kinetic Friction
Kinetic friction is the force that opposes the relative motion of two objects in contact with each other when they are moving past one another. It acts parallel to the surfaces in contact and is directed opposite to the direction of motion.The magnitude of kinetic friction \( f_k \) can be calculated using the formula:
  • \( f_k = \mu_k \cdot N \)
Where \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force.
In the given exercise, the coefficient of kinetic friction is \(0.35\), and we used the normal force of 539 N (calculated from the weight of the crate). This results in a kinetic frictional force of \(188.65 \text{ N}\). Understanding kinetic friction helps us to predict and calculate the resistance an object will face when being moved across a surface.
Normal Force
The normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface. For objects resting on a horizontal surface, the normal force is usually equal to the gravitational force (weight) of the object. This force can be calculated using the formula:
  • \( N = m \cdot g \)
Where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, typically \(9.8 \text{ m/s}^2\) on Earth.
In the exercise, the crate's weight is \(55 \text{ kg} \times 9.8 \text{ m/s}^2\), giving a normal force of \(539 \text{ N}\). Understanding the normal force is crucial as it helps in calculating other forces like friction, which are vital in analyzing motion scenarios.

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Most popular questions from this chapter

SSM WWW A bedroom bureau with a mass of \(45 \mathrm{~kg}\), including drawers and clothing, rests on the floor. (a) If the coefficient of static friction between the bureau and the floor is \(0.45\), what is the magnitude of the minimum horizontal force that a person must apply to start the bureau moving? (b) If the drawers and clothing, with \(17 \mathrm{~kg}\) mass, are removed before the bureau is pushed, what is the new minimum magnitude?

A student of weight 667 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force \(\vec{F}_{N}\) on the student from the seat is \(556 \mathrm{~N}\). (a) Does the student feel "light" or "heavy" there? (b) What is the magnitude of \(\vec{F}_{N}\) at the lowest point? If the wheel's speed is doubled, what is the magnitude \(F_{N}\) at the (c) highest and (d) lowest point?

A \(110 \mathrm{~g}\) hockey puck sent sliding over ice is stopped in \(15 \mathrm{~m}\) by the frictional force on it from the ice. (a) If its initial speed is \(6.0 \mathrm{~m} / \mathrm{s}\), what is the magnitude of the frictional force? (b) What is the coefficient of friction between the puck and the ice?

A locomotive accelerates a 25 -car train along a level track. Every car has a mass of \(5.0 \times 10^{4} \mathrm{~kg}\) and is subject to a frictional force \(f=250 v\), where the speed \(v\) is in meters per second and the force \(f\) is in newtons. At the instant when the speed of the train is 30 \(\mathrm{km} / \mathrm{h}\), the magnitude of its acceleration is \(0.20 \mathrm{~m} / \mathrm{s}^{2} .(\mathrm{a})\) What is the tension in the coupling between the first car and the locomotive? (b) If this tension is equal to the maximum force the locomotive can exert on the train, what is the steepest grade up which the locomotive can pull the train at \(30 \mathrm{~km} / \mathrm{h}\) ?

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is \(5.0 \mathrm{kN}\), and the circle's radius is \(10 \mathrm{~m}\). At the top of the circle, what are the (a) magnitude \(F_{B}\) and (b) direction (up or down) of the force on the car from the boom if the car's speed is \(v=5.0 \mathrm{~m} / \mathrm{s} ?\) What are (c) \(F_{B}\) and (d) the direction if \(v=12 \mathrm{~m} / \mathrm{s} ?\)

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