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Ch 05: Applying Newton's Laws
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 5, Problem 29a

A 45.045.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313313 N. Then you must reduce your push to 208208 N to keep it moving at a steady 25.025.0 cm/s. What are the coefficients of static and kinetic friction between the crate and the floor?

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Step 1: Understand the problem. The problem involves determining the coefficients of static and kinetic friction. The coefficient of static friction (μ_s) is related to the force required to start moving the crate, while the coefficient of kinetic friction (μ_k) is related to the force required to keep the crate moving at a constant velocity.
Step 2: Write the formula for the force of friction. The force of friction is given by the equation: f=μN, where μ is the coefficient of friction and N is the normal force. For a horizontal surface, the normal force is equal to the weight of the crate: N=mg, where m is the mass of the crate and g is the acceleration due to gravity.
Step 3: Calculate the coefficient of static friction. The force required to overcome static friction is given as 313 N. Using the formula for static friction: f=μN, substitute f=313 N and N=mg=45.0 kg 9.8 m/s². Solve for μs.
Step 4: Calculate the coefficient of kinetic friction. The force required to keep the crate moving at a constant velocity is given as 208 N. Using the formula for kinetic friction: f=μkN, substitute f=208 N and N=mg=45.0 kg 9.8 m/s². Solve for μk.
Step 5: Interpret the results. The coefficients of static and kinetic friction are dimensionless quantities that describe the interaction between the crate and the floor. A higher coefficient of static friction indicates that more force is required to start moving the crate, while a lower coefficient of kinetic friction indicates that less force is required to keep it moving at a constant velocity.

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Key Concepts

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

Static Friction

Static friction is the force that must be overcome to start moving an object at rest. It acts in the opposite direction of the applied force and varies up to a maximum value, which is determined by the coefficient of static friction and the normal force. In this scenario, the maximum static friction force is equal to the applied force just before the crate begins to move.
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Kinetic Friction

Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. It is generally less than static friction and is characterized by the coefficient of kinetic friction, which, when multiplied by the normal force, gives the frictional force acting on a moving object. In this case, the kinetic friction force is what allows the crate to move at a constant speed once it is in motion.
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Normal Force

The normal force is the perpendicular force exerted by a surface to support the weight of an object resting on it. It acts against gravity and is crucial for calculating frictional forces. In this problem, the normal force is equal to the weight of the crate, which is the product of its mass and the acceleration due to gravity, and it directly influences both static and kinetic friction.
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Related Practice
Textbook Question

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing. The toolbox will slide out if it is set moving. The coefficients of kinetic friction and static friction between the box and the level bed of the truck are 0.3550.355 and 0.6500.650, respectively. Starting from rest, what is the shortest time this truck could accelerate uniformly to 30.030.0 m/s without causing the box to slide? Draw a free-body diagram of the toolbox.

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Textbook Question

A box of bananas weighing 40.0 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.400.40, and the coefficient of kinetic friction is 0.200.20. What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started?

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Textbook Question

A box of bananas weighing 40.040.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.400.40, and the coefficient of kinetic friction is 0.200.20. If the monkey applies a horizontal force of 18.018.0 N, what is the magnitude of the friction force and what is the box's acceleration?

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Textbook Question

A box of bananas weighing 40.0 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.400.40, and the coefficient of kinetic friction is 0.200.20. What minimum horizontal force must the monkey apply to start the box in motion?

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Textbook Question

A 45.045.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313313 N. Then you must reduce your push to 208208 N to keep it moving at a steady 25.025.0 cm/s. Suppose you were performing the same experiment on the moon, where the acceleration due to gravity is 1.621.62 m/s2.

(i) What magnitude push would cause it to move?

(ii) What would its acceleration be if you maintained the push in part (b)? Note: Part (b) asked what push you must exert to give it an acceleration of 1.101.10 m/s2.

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Textbook Question

A 45.045.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313313 N. Then you must reduce your push to 208208 N to keep it moving at a steady 25.025.0 cm/s. What push must you exert to give it an acceleration of 1.101.10 m/s2?

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