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Matter & Interactions
Found in: Page 167
Matter & Interactions

Matter & Interactions

Book edition 4th edition
Author(s) Ruth W. Chabay, Bruce A. Sherwood
Pages 1135 pages
ISBN 9781118875865

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Short Answer

A 5 kg box with initial speed 4m/sslides across the floor and comes to a stop after 0.7s (a) what is the coefficient of kinetic friction? (b) How far does the box move? (c) You put a 3kg block in the box, so the total mass is now 8kg, and you launch this heavier box with an initial speed of 4m/s. How long does it take to stop?

  1. The coefficient of kinetic friction is 1.17

  2. The box moves 0.7 m

  3. The time taken to stop a box of 8kg is 0.56s

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Identification of the given data

The mass of a box is 5 kg.

The initial speed is 4m/s.

Time taken to stop the box is 0.7s.

Step 2: Definition of frictional force and the coefficient of kinetic friction

Frictional force is defined as the force required resisting the motion of a box when one object’s surface contacts with other ones object.

The coefficient of kinetic friction is defined as the ratio of kinetic friction force to normal force.

Step 3: (a) Determination of the coefficient of kinetic friction

Coefficient of kinetic friction,

μ=FN ..(1)

Where, F = Frictional Force

N = Normal Force

To find Frictional Force,

F = ma …(2)

Where, m = mass of a box

a = acceleration

To find acceleration a,

v2=u2+2as ..(3)

Where, v = 0

u = 4 m/s

s = 0.7 s

Substitute these values in Equation (3),

0=42+2×a×0.7 0=16+1.4a1.4a=-16 a=11.43 m/s2

Substitute “a” value in Equation (2),

F=maF=5×11.43F=57.14N

Frictional Force is 57.14N

To find Normal Force N,

N = mg …(4)

Where,

m= mass

g=9.8 m/s2

Substitute these values in Equation (4),

N=mgN=5×9.8N=49N

From Equation (1),

Coefficient of kinetic friction,

μ=FN =57.1449 =1.17

Hence, the Coefficient of kinetic friction is role="math" 1.17

Step 4: (b) Determination of the distance moved by the box

F×d=12mv2 …(5)

Where, d =distance

From the Equation (5),

F×d=12mv2d=12mv2F =12×5×4257.14 =0.7m

Hence, the box moves 0.7m.

Step 5: (c) Determination of the distance moved by the box

Time taken to stop the box,

F=m(v-u)tt=m(v-u)F =8(0-4)57.14 =0.56s

Hence, the Time taken to stop the box is 0.56s

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