Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781337026345
Author: Katz
Publisher: Cengage
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Chapter 5, Problem 74PQ

Starting from rest, a rectangular toy block with mass 300 g slides in 1.30 s all the way across a table 1.20 m in length that Zak has tilted at an angle of 42.0° to the horizontal. a. What is the magnitude of the acceleration of the toy block? b. What is the coefficient of kinetic friction between the block and the table? c. What are the magnitude and direction of the friction force acting on the block? d. What is the speed of the block when it is at the end of the table, having slid a distance of 1.20 m?

(a)

Expert Solution
Check Mark
To determine

Find the magnitude of acceleration of the toy block.

Answer to Problem 74PQ

The magnitude of acceleration of the toy block is 1.42 m/s2.

Explanation of Solution

Write the equation for acceleration.

  x=12at2a=2xt2                                                                                 (I)

Here, a is the acceleration, x is the displacement and t is the time.

Conclusion:

Substitute 1.20 m for x and 1.30 s for t in equation I.

    a=2(1.20 m)(1.30 s)2=1.42 m/s2

Therefore, the magnitude of acceleration of the toy block is 1.42 m/s2.

(b)

Expert Solution
Check Mark
To determine

Find the coefficient of kinetic friction between the block and table.

Answer to Problem 74PQ

The coefficient of kinetic friction between the block and table is 0.706.

Explanation of Solution

From the given condition the free body diagram is given below.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 5, Problem 74PQ

  ΣFx=Ffmgsinθ=maFf=m(gsinθa)                                                                                (II)

Here, Fx is the force, Ff is the friction force, m is the mass, a is the acceleration and θ is the angle.

  ΣFy=FNmgcosθ=0FN=mgcosθ                                                                       (III)

Here, FN is the normal force.

Write the equation for friction force.

    Ff=μkFN                                                                                             (IV)

Here, μk is the coefficient of friction.

Substitute equation II and III in equation IV.

    μk=m(gsinθa)mgcosθ=tanθagcosθ                                                                                (V)

Conclusion:

Substitute 42.0° for θ, 9.81 m/s2 for g and 1.42 m/s2 for a in equation V.

    μk=tan(42.0°)1.42 m/s2(9.81 m/s2)cos(42.0°)=0.706

Therefore, the coefficient of kinetic friction between the block and table is 0.706.

(c)

Expert Solution
Check Mark
To determine

Find the direction and magnitude of friction acting on the block.

Answer to Problem 74PQ

The friction acting on the block is directed upward and its magnitude is 1.54 N.

Explanation of Solution

Substitute equation III in IV.

  Ff=μkmgcosθ                                                                       (VI)

Conclusion:

Substitute 42.0° for θ, 9.81 m/s2 for g, 1.42 m/s2 for a, 0.706 for μk and 0.300 kg for m  in equation VI.

    Ff=(0.706)(0.300 kg)(9.81 m/s2)cos(42.0°)=1.54 N

The friction force is directed up in the incline.

Therefore, the friction acting on the block is directed upward and its magnitude is 1.54 N.

(d)

Expert Solution
Check Mark
To determine

Find the speed of the block when it is at the end of the table.

Answer to Problem 74PQ

The speed of the block when it is at the end of the table is 1.85 m/s.

Explanation of Solution

Write the equation of motion.

  vf2=vi2+2aΔx                                                                       (VII)

Here, vf is the final speed, vi is the initial speed, Δx is the displacement and a is the acceleration.

Conclusion:

Substitute 0 for vi, 1.42 m/s2 for a and 1.2 m for Δx in equation VII.

    vf=0+2(1.42 m/s2)(1.2 m)=1.85 m/s

Therefore, the speed of the block when it is at the end of the table is 1.85 m/s.

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Chapter 5 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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