Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 5, Problem 5.22P

Review. Three forces acting on an object are given by F 1 = ( 2.00 i ^ 2.00 j ^ ) N , and F 1 = ( 5.00 i ^ 3.00 j ^ ) N , and F 1 = ( 45.0 i ^ ) N . The object experiences an acceleration of magnitude 3.75 m/s2. (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after 10.0 s?

(a)

Expert Solution
Check Mark
To determine

The direction of the acceleration.

Answer to Problem 5.22P

The direction of the acceleration is 181°.

Explanation of Solution

The forces acting on the object are F1=(2.00i^+2.00j^)N, F2=(5.00i^3.00j^)N and F3=(45.00i^)N. The magnitude of acceleration of the object is 3.75m/s2.

Write the formula to calculate net force act on a object

    Fnet=F1+F2+F3

Here, Fnet is the net force acting on an object, F1, F2 and are the given forces.

Substitute (2.00i^+2.00j^)N for F1, (5.00i^3.00j^)N for F2 and (45.00i^)N for F3 to find Fnet.

    Fnet=(2.00i^+2.00j^)N+(5.00i^3.00j^)N+(45.00i^)N=(42.0i^1.00j^)N

Write the formula to calculate direction of force

    tanθ=(FyFx)

Here, Fy is the y-component of the force and Fx is the x-component of the force.

Conclusion:

Substitute 42.0N for Fx and 1.00N for Fy in the above equation to calculate θ.

    tanθ=(1.00N42.0N)θ=1.36°

The direction of force is equal to the direction of acceleration of object and the value of θ lies at third quadrant so the direction of acceleration with +x axis

    θ=180°+1.36°=181.36°181°

Therefore, the direction of the acceleration is 181°.

(b)

Expert Solution
Check Mark
To determine

The mass of the object.

Answer to Problem 5.22P

The mass of the object is 11.2kg.

Explanation of Solution

Write the formula to calculate magnitude of net force act of an object

    F=Fx2+Fy2

Write the formula to calculate mass of the object

    m=Fa

Here, a is the magnitude of acceleration of an object, m is the mass of the object and F is the magnitude of net force acting on the object.

Substitute Fx2+Fy2 for F in the above equation.

    m=Fx2+Fy2a

Conclusion:

Substitute 42.0N for Fx and 1.00N for Fy  and 3.75m/s2 for a in the above equation to find m.

    m=(42.0N)2+(1.00)23.75m/s2=11.2kg

Therefore, the mass of the object is 11.2kg.

(c)

Expert Solution
Check Mark
To determine

The speed of the object after 10sec.

Answer to Problem 5.22P

The speed of the object after 10sec is 37.5m/s.

Explanation of Solution

Write the formula to calculate speed of an object

    vf=vi+at

Here, vf is the final speed of an object, vi is the initial speed of an object and t is time.

Conclusion:

Substitute 0 for vi, 10sec for t and 3.75m/s2 for a to find vf.

    vf=0+3.75m/s2×10sec=37.5m/s

Therefore, the speed of the object after 10sec is 37.5m/s.

(d)

Expert Solution
Check Mark
To determine

The velocity components of the object after 10sec.

Answer to Problem 5.22P

The x and y components of the velocity are 37.5m/s and 0.893m/s respectively.

Explanation of Solution

Write the formula to calculate velocity of an object

    vf=vi+at                          (I)

Here, vf is the final velocity and vi is the initial velocity.

Write the formula to calculate mass of the object

    a=Fnetm

Substitute Fnetm for a in equation (I).

    vf=vi+Fnetmt

Substitute (42.0i^1.00j^)N for Fnet, 11.2kg for m, 0 for vi and 10sec for t to find vf.

    vf=0+(42.0i^1.00j^)N11.2kg10sec=(37.5i^0.893j^)m/s

Conclusion:

Therefore, x and y components of the velocity are 37.5m/s and 0.893m/s respectively.

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Chapter 5 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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