A box of bananas weighing 40.0 40.040.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.400.400.40, and the coefficient of kinetic friction is 0.200.200.20. What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started?

Ch 05: Applying Newton's Laws

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 05: Applying Newton's Laws

Problem 28d

Chapter 5, Problem 28d

A box of bananas weighing $40.0$ N rests on a horizontal surface. The coefficient of static friction between the box and the surface is $0.40$ , and the coefficient of kinetic friction is $0.20$ . What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started?

Verified step by step guidance

Step 1: Understand the problem. The box is already moving, and the goal is to find the minimum horizontal force required to keep it moving at a constant velocity. This means the applied force must balance the kinetic friction force acting on the box.

Step 2: Recall the formula for the force of kinetic friction: $ F_k = \mu_k \cdot F_N $, where $ \mu_k $ is the coefficient of kinetic friction and $ F_N $ is the normal force. Here, $ \mu_k = 0.20 $ and $ F_N $ is equal to the weight of the box, $ F_N = 40.0 \, \text{N} $.

Step 3: Substitute the given values into the formula for $ F_k $: $ F_k = 0.20 \cdot 40.0 \, \text{N} $. This will give the force of kinetic friction acting on the box.

Step 4: Since the box is moving at a constant velocity, the applied force $ F_{\text{applied}} $ must equal the force of kinetic friction $ F_k $. Therefore, $ F_{\text{applied}} = F_k $.

Step 5: Conclude that the minimum horizontal force required to keep the box moving at constant velocity is equal to the calculated kinetic friction force. Perform the calculation to find the numerical value if needed.

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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 between the surfaces in contact and is dependent on the normal force and the coefficient of static friction. In this case, the static friction force can be calculated using the formula: F_static = μ_static * N, where μ_static is the coefficient of static friction and N is the normal force.

Kinetic Friction

Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. Once the box of bananas is in motion, the kinetic friction force comes into play, which is generally less than static friction. It can be calculated using the formula: F_kinetic = μ_kinetic * N, where μ_kinetic is the coefficient of kinetic friction.

Newton's First Law of Motion

Newton's First Law states that an object at rest will remain at rest, and an object in motion will continue in motion at a constant velocity unless acted upon by a net external force. In this scenario, to keep the box moving at constant velocity, the applied force must equal the kinetic friction force, ensuring that the net force is zero.

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Textbook Question

A box of bananas weighing $40.0$ N rests on a horizontal surface. The coefficient of static friction between the box and the surface is $0.40$ , and the coefficient of kinetic friction is $0.20$ . If the monkey applies a horizontal force of $18.0$ 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 $40.0$ N rests on a horizontal surface. The coefficient of static friction between the box and the surface is $0.40$ , and the coefficient of kinetic friction is $0.20$ . What minimum horizontal force must the monkey apply to start the box in motion?

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Textbook Question

A $45.0$ -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 $313$ N. Then you must reduce your push to $208$ N to keep it moving at a steady $25.0$ cm/s. What push must you exert to give it an acceleration of $1.10$ m/s²?

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Textbook Question

A box of bananas weighing $40.0$ N rests on a horizontal surface. The coefficient of static friction between the box and the surface is $0.40$ , and the coefficient of kinetic friction is $0.20$ . If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it?

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Textbook Question

A $45.0$ -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 $313$ N. Then you must reduce your push to $208$ N to keep it moving at a steady $25.0$ cm/s. What are the coefficients of static and kinetic friction between the crate and the floor?

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Textbook Question

A box of bananas weighing $40.0$ N rests on a horizontal surface. The coefficient of static friction between the box and the surface is $0.40$ , and the coefficient of kinetic friction is $0.20$ . What is the magnitude of the friction force if a monkey applies a horizontal force of $6.0$ N to the box and the box is initially at rest?

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