Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 4, Problem 4P

Two forces, F 1 = ( 6 i ^ 4 j ^ ) N and F 2 = ( 3 i ^ + 7 j ^ ) N , act on a particle of mass 2.00 kg that is initially at rest at coordinates (−2.00 m, +4.00 m). (a) What are the components of the particle’s velocity at t = 10.0 s? (b) In what direction is the particle moving at t = 10.0 s? (c) What displacement does the particle undergo during the first 10.0 s? (d) What are the coordinates of the particle at t = 10.0 s?

(a)

Expert Solution
Check Mark
To determine

The components of the particle’s velocity.

Answer to Problem 4P

The components of the particle’s velocity is 45.0m/s , 15.0 m/s_.

Explanation of Solution

Write the expression for velocity.

v=at        (I)

Here, v is the velocity, a is the acceleration and t is the time.

Write the expression for acceleration.

a=Fnetm        (II)

Here, Fnet is the total force, m is the mass.

Write the expression for net force.

Fnet=F1+F2        (III)

Here, F1,F2 are the applied forces.

Rewrite the expression for (I) by using (II) and (III).

v=(F1+F2m)t        (IV)

Conclusion:

Substitute, (6i^4j^)N for F1, (3i^+7j^)N for F2, 2.00kg for m, 10.0s for t in equation (IV) to find v.

v=(((6i^4j^)N)+((3i^+7j^)N)(2.00kg))(10.0s)=[(4.50i^+1.50j^)m/s2](10.0s)=(45.0i^+15j^)m/s

Thus, the components of the particle’s velocity is 45.0m/s , 15.0 m/s_.

(b)

Expert Solution
Check Mark
To determine

The direction of the moving particle.

Answer to Problem 4P

The direction of the moving particle is 162° from the x-axis_.

Explanation of Solution

Write the expression for the direction of motion.

θ=180°+[tan1(vyvx)]        (V)

Here, θ is the angle, vx,vy are the x and y component of velocity.

Conclusion:

Substitute, 15.0m/s for vy, 45.0m/s for vx in equation (V) to find θ.

θ=180°+[tan1(15.0m/s45.0m/s)]=180°+(18.4°)162°from the x-axis

Thus, the direction of the moving particle is 162° from the x-axis_.

(c)

Expert Solution
Check Mark
To determine

The displacement of the particle.

Answer to Problem 4P

The displacement of the particle is (225i^+75.0j^)m_.

Explanation of Solution

Write the expression for displacement.

(Δr)=(Δx)+(Δy)        (IV)

Here, (Δr) is the displacement, (Δx) is the x-displacement, (Δy) is the y-displacement.

Write the expression for x-displacement.

(Δx)=vxit+12(axi^)t2        (V)

Here, vxi is the initial velocity along x-axis, t is the time, axi^ is the x-component of acceleration.

Write the expression for y-displacement.

(Δy)=vyit+12(ayj^)t2        (VI)

Here, vyi is the initial velocity along y-axis, t is the time, ayi^ is the y-component of acceleration.

Write the expression for the acceleration.

a=axi^+ayj^        (VII)

Rewrite the expression for the displacement by using (II), (III), (V), (VI) and (VII).

(Δr)=[vxit+12(axi^)t2]+[vyit+12(ayj^)t2]=(vxi+vyi)t+12t2(axi^+ayj^)=(vxi+vyi)t+12t2a=(vxi+vyi)t+12t2(F1+F2m)        (VIII)

Conclusion:

Substitute, (6i^4j^)N for F1, (3i^+7j^)N for F2, 2.00kg for m, 10.0s for t, 0 for vxi, 0 for vyi in equation (VIII) to find (Δr).

(Δr)=[(0+0)(10.0s)]+[12(10.0s)2([(6i^4j^)N]+[(3i^+7j^)N](2.00kg))]=[12(10.0s)2([(6i^4j^)N]+[(3i^+7j^)N](2.00kg))]=(225i^+75.0j^)m

Thus, the displacement of the particle is (225i^+75.0j^)m_.

(d)

Expert Solution
Check Mark
To determine

The coordinates of the particle.

Answer to Problem 4P

The coordinates of the particle is (227m , 79.0m)_.

Explanation of Solution

Write the expression for position of the particle.

rf=ri+(Δr)        (IX)

Here, rf is the final position, ri is the initial position.

Conclusion:

Substitute, (2.00i^+4.00j^)m for ri, (225i^+75j^)m for (Δr) in equation (IX) to find rf.

rf=((2.00i^+4.00j^)m)+((225i^+75.0j^)m)=(227i^+79.0j^)m

Thus, the coordinates of the particle is (227m , 79.0m)_.

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Chapter 4 Solutions

Principles of Physics: A Calculus-Based Text

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