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?

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

All textbooks

Young & Freedman Calc 14th Edition

Ch 05: Applying Newton's Laws

Problem 28e

Chapter 5, Problem 28e

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?

Verified step by step guidance

Step 1: Calculate the maximum static friction force using the formula:

f s_{\max} = μ s N

, where

μ s

is the coefficient of static friction and

N

is the normal force. Here,

N

equals the weight of the box, which is 40.0 N. Substitute

μ s = 0.40

and

N = 40.0

N into the formula.

Step 2: Compare the applied force (18.0 N) to the maximum static friction force calculated in Step 1. If the applied force is less than or equal to the maximum static friction force, the box does not move, and the friction force equals the applied force. If the applied force exceeds the maximum static friction force, the box starts moving, and kinetic friction comes into play.

Step 3: If the box moves, calculate the kinetic friction force using the formula:

f k = μ k N

, where

μ k

is the coefficient of kinetic friction. Substitute

μ k = 0.20

and

N = 40.0

N into the formula.

Step 4: If the box moves, calculate the net force acting on the box using the formula:

F net = F applied - f k

, where

F applied

is the applied force and

f k

is the kinetic friction force.

Step 5: Use Newton's second law to calculate the acceleration of the box if it moves. The formula is:

a = \frac{F}{net}

, where

m

is the mass of the box. To find the mass, use the relation

m = \frac{W}{g}

, where

W

is the weight of the box (40.0 N) and

g

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

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Static and Kinetic Friction

Static friction is the force that prevents an object from starting to move when a force is applied. It is characterized by a coefficient of static friction, which in this case is 0.40. Kinetic friction, on the other hand, acts on an object that is already in motion, with a lower coefficient of kinetic friction (0.20). Understanding these concepts is crucial for determining the frictional forces acting on the box when the monkey applies a force.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma). In this scenario, the net force is the difference between the applied force and the frictional force. This law is essential for calculating the box's acceleration once the frictional force is determined.

Calculating Frictional Force

The frictional force can be calculated using the formula F_friction = μ * N, where μ is the coefficient of friction and N is the normal force. For the box resting on a horizontal surface, the normal force equals its weight (40.0 N). Depending on whether the box is stationary or moving, the appropriate coefficient (static or kinetic) must be used to find the frictional force, which is critical for solving the problem.

Recommended video:

Guided course

04:29

Work Done By Friction

Related Practice

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 keep the box moving at constant velocity once it has been started?

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. Suppose you were performing the same experiment on the moon, where the acceleration due to gravity is $1.62$ 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 $1.10$ m/s².

views

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?

views

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?

views

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?

views