Part (a) Answer Solution : 64.7 N Explanation Given: Mass of the crate, m = 22 kg Coefficient of kinetic friction, μ k = 0.30 Formula used : Frictional force F is obtained by: F = μ N Where, µ is the coefficient of friction and N is the normal reaction force. The normal force would be equal to the weight of the crate. N = m g F = µ k m g Where F is the force acting, m is the mass, g is the acceleration due to gravity, µ k is the coefficient of kinetic friction between the crate and the floor. Calculation : The crate is required to be moved at steady speed which means acceleration is zero. The required horizontal force would be just equal to the kinetic frictional force. F = µ k m g F = ( 0.30 ) ( 22 kg ) ( 9.8 m/s 2 ) F = 64. 7 N Conclusion : The horizontal force is 64.7 N. Part (b) To determine The horizontal force if µ k is zero. Answer Solution : 0 N Explanation Given: Mass of the crate, m = 22 kg Coefficient of kinetic friction, μ k = 0 Formula used : Frictional force F is obtained by: F = μ N Where, µ is the coefficient of friction and N is the normal reaction force. The normal force would be equal to the weight of the crate. N = m g F = µ k m g Where F is the force acting, m is the mass, g is the acceleration due to gravity, µ k is the coefficient of kinetic friction between the crate and the floor. Calculation : The crate is required to be moved at steady speed which means acceleration is zero. The required horizontal force would be just equal to the kinetic frictional force. F = µ k m g = ( 0 ) m g = 0 N Conclusion : Therefore, the force required to move the crate on a frictionless surface is zero.

6th Edition

ISBN: 9780130606204

Author: Douglas C. Giancoli

Publisher: Prentice Hall

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Newton's Second Law

Newton’s laws of motion describe the relationship between the motion of an object and forces acting on it. Newton’s law of motion has two types as follows:

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Question

Chapter 4, Problem 34P

Part (a)

To determine

The horizontal force required to move the crate at a steady speed across the floor.

Part (a)

Expert Solution

Answer to Problem 34P

Solution:

64.7 N

Explanation of Solution

Given:

Mass of the crate, $m=22 kg$

Coefficient of kinetic friction, $μk=0.30$

Formula used:

Frictional force F is obtained by:

$F=μN$

Where, $µ$ is the coefficient of friction and N is the normal reaction force.

The normal force would be equal to the weight of the crate.

$N=mg$

$F=µkmg$

Where F is the force acting, m is the mass, g is the acceleration due to gravity, $µk$ is the coefficient of kinetic friction between the crate and the floor.

Calculation:

The crate is required to be moved at steady speed which means acceleration is zero. The required horizontal force would be just equal to the kinetic frictional force.

$F=µkmg$

$F= (0.30) (22 kg) (9.8 m/s2)$

$F=64.7 N$

Conclusion:

The horizontal force is 64.7 N.

Part (b)

To determine

The horizontal force if $µk$ is zero.

Part (b)

Expert Solution

Answer to Problem 34P

Solution:

0 N

Explanation of Solution

Given:

Mass of the crate, $m=22 kg$

Coefficient of kinetic friction, $μk=0$

Formula used:

Frictional force F is obtained by:

$F=μN$

Where, $µ$ is the coefficient of friction and N is the normal reaction force.

The normal force would be equal to the weight of the crate.

$N=mg$

$F=µkmg$

Where F is the force acting, m is the mass, g is the acceleration due to gravity, $µk$ is the coefficient of kinetic friction between the crate and the floor.

Calculation:

The crate is required to be moved at steady speed which means acceleration is zero. The required horizontal force would be just equal to the kinetic frictional force.

$F=µkmg$

$= (0) mg$

$=0 N$

Conclusion:

Therefore, the force required to move the crate on a frictionless surface is zero.

Chapter 4, Problem 33P

Chapter 4, Problem 35P

Chapter 4 Solutions

Physics: Principles with Applications

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