Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Figure 6-20 shows an initially stationary block of mass on a floor. A force of magnitude is then applied at upward angle $θ = 20 °$ .What is the magnitude of the acceleration of the block across the floor if the friction coefficients are (a) $μ s = 0.600$ and $μ_{k} = 0.500$ and (b) $μ s = 0.400$ and $μ_{k} = 0.300$ ?

(a) The magnitude of the acceleration of the block across the floor is 0 . $\frac{m}{s^{2}}$

(b) The magnitude of the acceleration of the block across the floor is 2.15. $\frac{m}{s^{2}}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

Mass of the block, m

Force on the block, $F = 0.500 m g . N$

Angle, $θ = 20^{0}$

$μ_{s} =$ 0.600 and $μ_{K} =$ 0.500

$μ_{s} =$ 0.400 and $μ_{K} =$ 0.300

Step 2: Determining the concept and the formula:

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motionalong vertical and horizontal direction.

Write the formula for the net force:

$F_{n e t} = \sum m a$

Here, F is the net force, m is mass and a is an acceleration.

Step 3: (a) Determinethe magnitude of the acceleration of the block across the floor if the friction coefficients are μs =0.600 and μk =0.500

By using Newton’s 2nd law along y direction,block is not moving along y. So, the acceleration is 0.

$\sum_{}^{} F_{y} = 0$

$N + F \sin (20) - m g = 0$

Substitute the values and solve as:

$N = m g - (0.500 mg) \sin (20) = 0.83 m g N$

Static frictional force on the block from the floor is,

$\begin{array}{rcl} f_{s} & = & μ_{s} N \\ = & (0.600) (0.83 mg) \\ = & 0.50 mgN \end{array}$

Opposite force to the $f_{s}$ is,

$\begin{array}{rcl} F \cos (20) & = & (0.500 mg) \cos (20) \\ = & 0.47 mg \cdot N \end{array}$

So, here, $F \cos (20) < f_{s}$ therefore block cannot move.

Therefore, the magnitude of the acceleration of the block across the floor is 0 . $\frac{m}{s^{2}}$

Step 4: (b) Determining the magnitude of the acceleration of the block across the floor if the friction coefficients are μs =0.400andμk=0.300

Static frictional force on the block from the floor is,

$\begin{array}{rcl} f_{s} & = & μ_{s} N \\ = & (0.400) (0.83 mg) \\ = & 0.33 mg \cdot N \end{array}$

Opposite force to the $f_{s}$ is,

$\begin{array}{rcl} F \cos (20) & = & (0.500 m g) \cos (20) \\ = & 0.47 m g . N \end{array}$

So,here . $F \cos (20) > f_{s}$ Therefore, the block will be moved.

Kinetic frictional force on the block is,

$\begin{array}{rcl} f_{k} & = & μ_{k} N \\ = & (0.300) (0.83 mg) \\ = & 0.25 m g . N \end{array}$

By using Newton’s 2nd law along x direction is,

$\sum_{}^{} F_{x} = m a_{x} F \cos (20) - f_{k} = m a_{x} (0.500 m g) \cos (20) - (0.25 m g) = m a_{x} a_{x} = 2.15 \frac{m}{s^{2}}$

Therefore, the magnitude of the acceleration of the block across the floor is 2.15. $\frac{m}{s^{2}}$

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Short Answer

Step by Step Solution

Step 1: Given

Step 2: Determining the concept and the formula:

Step 3: (a) Determinethe magnitude of the acceleration of the block across the floor if the friction coefficients are μs =0.600 and μk =0.500

Step 4: (b) Determining the magnitude of the acceleration of the block across the floor if the friction coefficients are μs =0.400andμk=0.300

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Fundamentals Of Physics

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Short Answer

Figure 6-20 shows an initially stationary block of mass on a floor. A force of magnitude is then applied at upward angleθ =20° .What is the magnitude of the acceleration of the block across the floor if the friction coefficients are (a) μs =0.600andμk=0.500 and (b)μs =0.400 andμk=0.300 ?

Step by Step Solution

Step 1: Given

Step 2: Determining the concept and the formula:

Step 3: (a) Determinethe magnitude of the acceleration of the block across the floor if the friction coefficients are μs =0.600 and μk =0.500

Step 4: (b) Determining the magnitude of the acceleration of the block across the floor if the friction coefficients are μs =0.400andμk=0.300

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