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 3e

Chapter 6, Problem 3e

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?

Verified step by step guidance

Step 1: Identify the forces acting on the crate. Since the crate is moving at constant velocity, the net force is zero. The forces include the applied horizontal force, the kinetic friction force, and the normal force. The kinetic friction force can be calculated using the formula:

f

_k=μ_kN, where

μ

_k is the coefficient of kinetic friction and

N

is the normal force.

Step 2: Calculate the normal force. Since the floor is level, the normal force is equal in magnitude to the gravitational force acting on the crate. Use the formula:

N = m g

, where

m

is the mass of the crate and

g

is the acceleration due to gravity (approximately 9.8 m/s²).

Step 3: Calculate the kinetic friction force using the formula:

f

_k=μ_kN. Substitute the values for

μ

_k and

N

from the previous step.

Step 4: Determine the work done by the worker to overcome friction. The work done is given by the formula:

W = f

_kd, where

d

is the distance the crate is pushed. Substitute the values for

f

_k and

d

Step 5: Since the crate is moving at constant velocity, the total work done on the crate is equal to the work done to overcome friction. This is because there is no change in kinetic energy (no net work done). Use the value calculated in the previous step as the total work done.

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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 crate is pushed horizontally and moves in the same direction, the angle θ is 0 degrees, simplifying the calculation.

Friction

Friction is a force that opposes the relative motion of two surfaces in contact. The coefficient of kinetic friction (μ_k) quantifies this force, calculated as F_friction = μ_k × N, where N is the normal force. In this case, the kinetic friction acts against the motion of the crate, and understanding its role is crucial for determining the net work done on the crate.

Net Work

Net work is the total work done on an object, taking into account all forces acting on it. It is calculated by subtracting the work done against friction from the work done by the applied force. Since the crate moves at a constant velocity, the net work done on it is zero, indicating that the work done by the worker is equal to the work done against friction.

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Calculating Net Work

Related Practice

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 $B$ ?

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Textbook Question

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?

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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 $0.25$ . How much work is done on the crate by this force?

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Textbook Question

You throw a $3.00$ -N rock vertically into the air from ground level. You observe that when it is $15.0$ m above the ground, it is traveling at $25.0$ m/s upward. Use the work–energy theorem to find the rock's speed just as it left the ground.

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. What is the total work done on the crate?

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Key Concepts

Work

Friction

Net Work

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?