Ch 06: Work & Kinetic Energy

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 06: Work & Kinetic Energy

Problem 3d

Chapter 6, Problem 3d

A factory worker pushes a $30.0$ -kg crate a distance of $4.5$ m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is $0.25$ . How much work is done on the crate by the normal force? By gravity?

Verified step by step guidance

Step 1: Understand the problem. The crate is moving at constant velocity, which means the net force acting on it is zero. The work done by the normal force and gravity needs to be calculated. Recall that work is given by the formula:

W = F d \cos (θ)

, where

F

is the force,

d

is the displacement, and

θ

is the angle between the force and displacement.

Step 2: Analyze the direction of the forces. The normal force acts perpendicular to the surface (upward), and gravity acts vertically downward. Since the crate is moving horizontally, the angle between the displacement and both the normal force and gravity is 90 degrees.

Step 3: Apply the work formula for the normal force. Substitute

θ = 90

degrees into the formula. The cosine of 90 degrees is zero, so the work done by the normal force is:

W = F d \cos (90) = 0

. This means the normal force does no work on the crate.

Step 4: Apply the work formula for gravity. Similarly, the angle between the displacement and the gravitational force is also 90 degrees. Using the same reasoning as in Step 3, the work done by gravity is:

W = F d \cos (90) = 0

. Gravity does no work on the crate.

Step 5: Conclude the solution. Since both the normal force and gravity act perpendicular to the direction of displacement, neither force contributes to the work done on the crate. The work done by the normal force and gravity is zero.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Work

Work is defined as the product of the force applied to an object and the distance over which that force is applied, in the direction of the force. Mathematically, it is expressed as W = F × d × cos(θ), where θ is the angle between the force and the direction of motion. In this scenario, since the normal force and gravitational force act perpendicular to the direction of motion, the work done by these forces is zero.

Normal Force

The normal force is the force exerted by a surface to support the weight of an object resting on it, acting perpendicular to the surface. In this case, the normal force balances the weight of the crate, which is equal to the gravitational force acting on it. Since the crate is on a level floor and not accelerating vertically, the normal force equals the weight of the crate.

Friction

Friction is the resistive force that opposes the motion of an object in contact with a surface. The coefficient of kinetic friction quantifies this resistance, indicating how much frictional force acts when the object is sliding. In this problem, while friction affects the horizontal motion of the crate, it does not contribute to the work done by the normal force or gravity, as these forces act vertically.

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Related Practice

Textbook Question

A factory worker pushes a $30.0$ -kg crate a distance of $4.5$ m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is $0.25$ . What is the total work done on the crate?

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Textbook Question

A $1.50$ -kg book is sliding along a rough horizontal surface. At point $A$ it is moving at $3.21$ m/s, and at point $B$ it has slowed to $1.25$ m/s. How much work was done on the book between $A$ and $B$ ?

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of $1.80 \times 10^{6}$ N, one $14 degrees$ west of north and the other $14 degrees$ east of north, as they pull the tanker $0.75$ km toward the north. What is the total work they do on the supertanker?

A factory worker pushes a 30.030.030.0-kg crate a distance of 4.54.54.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.250.250.25. How much work is done on the crate by the normal force? By gravity?

Verified video answer for a similar problem:

Key Concepts

Work

Normal Force

Friction

A factory worker pushes a $30.0$ -kg crate a distance of $4.5$ m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is $0.25$ . How much work is done on the crate by the normal force? By gravity?