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Ch 06: Work & Kinetic Energy
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 6, Problem 3c

A factory worker pushes a 30.030.0-kg crate a distance of 4.54.5 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 0.250.25. How much work is done on the crate by friction?

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Step 1: Identify the forces acting on the crate. The crate is moving at constant velocity, which means the net force is zero. The force of kinetic friction opposes the motion, and the worker's applied force balances it.
Step 2: Calculate the force of kinetic friction using the formula: fk=μkN, where μk is the coefficient of kinetic friction (0.25) and N is the normal force. Since the floor is level, the normal force equals the weight of the crate: N=mg, where m is the mass of the crate (30.0 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).
Step 3: Substitute the values into the formula for the force of kinetic friction: fk=μkmg. This gives the magnitude of the frictional force.
Step 4: Use the work formula to calculate the work done by friction: Wfriction=-fkd, where d is the distance the crate is pushed (4.5 m). The negative sign indicates that friction opposes the motion.
Step 5: Substitute the values for fk and d into the work formula to find the work done by friction. Ensure the units are consistent throughout the calculation.

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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, the work done by friction is negative since it opposes the motion of the crate.
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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 when objects are sliding past each other. In this case, the frictional force can be calculated using F_friction = μ_k × N, where N is the normal force, which equals the weight of the crate on a level surface.
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Constant Velocity

Constant velocity implies that an object moves at a steady speed in a straight line, meaning that the net force acting on it is zero. In this scenario, the worker's applied force must equal the frictional force to maintain constant velocity, indicating that the work done by the worker is balanced by the work done against friction.
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Related Practice
Textbook Question

A factory worker pushes a 30.030.0-kg crate a distance of 4.54.5 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 0.250.25. What is the total work done on the crate?

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Textbook Question

A factory worker pushes a 30.030.0-kg crate a distance of 4.54.5 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 0.250.25. What magnitude of force must the worker apply?

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Textbook Question

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80×1061.80\(\times\)10^6 N, one 14°14° west of north and the other 14°14° east of north, as they pull the tanker 0.750.75 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 30.030.0-kg crate a distance of 4.54.5 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 0.250.25. How much work is done on the crate by this force?

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Textbook Question

A factory worker pushes a 30.030.0-kg crate a distance of 4.54.5 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 0.250.25. How much work is done on the crate by the normal force? By gravity?

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