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 8, Problem 58P

(a)

To determine

The final velocity of the object.

(a)

Expert Solution
Check Mark

Answer to Problem 58P

The final velocity of the object is (20i^+7j^)m/s .

Explanation of Solution

Given information: The mass of the object is 3kg , the velocity of the object is 7j^m/s and the net force acting on the object is 12i^N . The time duration is 5s .

Write the expression of impulse momentum equation.

m(vfvi)=FΔt (1)

Here,

m is the mass of the object.

vf is the final velocity of the object.

vi is the initial velocity of the object.

F is the net force acting on the object.

Δt is the time duration.

Substitute 3kg for m , 7j^m/s for vi , 12i^N for F and 5s for Δt in equation (1) to find vf .

3kg(vf7j^m/s)=12i^N×5svf=(20i^+7j^)m/s

Thus, the final velocity of the object is (20i^+7j^)m/s .

Conclusion:

Therefore, the final velocity of the object is (20i^+7j^)m/s .

(b)

To determine

The acceleration of the object.

(b)

Expert Solution
Check Mark

Answer to Problem 58P

The acceleration of the object is 4i^m/s2 .

Explanation of Solution

Given information: The mass of the object is 3kg , the velocity of the object is 7j^m/s and the net force acting on the object is 12i^N . The time duration is 5s .

Write the expression to calculate the acceleration of the object.

a=(vfvi)Δt (2)

Here,

a is the acceleration of the object.

Substitute 7j^m/s for vi , (20i^+7j^)m/s for vf and 5s for Δt in equation (2) to find a .

a=(20i^+7j^)m/s7j^m/s5s=4i^m/s2

Thus, the acceleration of the object is 4i^m/s2 .

Conclusion:

Therefore, the acceleration of the object is 4i^m/s2 .

(c)

To determine

The acceleration of the object.

(c)

Expert Solution
Check Mark

Answer to Problem 58P

The acceleration of the object is 4i^m/s2 .

Explanation of Solution

Given information: The mass of the object is 3kg , the velocity of the object is 7j^m/s and the net force acting on the object is 12i^N . The time duration is 5s .

Write the expression to calculate the acceleration of the object.

a=Fm (3)

Substitute 12i^N for F and 3kg for m in equation (3) to find a .

a=12i^N3kg=4i^m/s2

Thus, the acceleration of the object is 4i^m/s2 .

Conclusion:

Therefore, the acceleration of the object is 4i^m/s2 .

(d)

To determine

The vector displacement of the object.

(d)

Expert Solution
Check Mark

Answer to Problem 58P

The vector displacement of the object is (50i^+35j^)m .

Explanation of Solution

Given information: The mass of the object is 3kg , the velocity of the object is 7j^m/s and the net force acting on the object is 12i^N . The time duration is 5s .

Write the expression to calculate the vector displacement of the object.

r=vit+12at2 (4)

Here,

r is the vector displacement of the object.

Substitute 7j^m/s for vi , 4i^m/s2 for a and 5s for t in equation (4) to find r .

r=7j^m/s×5s+12×4i^m/s2×(5s)2=(50i^+35j^)m

Thus, the vector displacement of the object is (50i^+35j^)m .

Conclusion:

Therefore, the vector displacement of the object is (50i^+35j^)m .

(e)

To determine

The work done on the object.

(e)

Expert Solution
Check Mark

Answer to Problem 58P

The work done on the object is 600J .

Explanation of Solution

Given information: The mass of the object is 3kg , the velocity of the object is 7j^m/s and the net force acting on the object is 12i^N . The time duration is 5s .

Write the expression to calculate the work done on the object.

W=FΔr . (5)

Here,

W is the work done on the object.

Substitute 12i^N for F and (50i^+35j^)m for Δr in equation (5) to find W .

W=12i^N(50i^+35j^)m=600J

Thus, the work done on the object is 600J .

Conclusion:

Therefore, the work done on the object is 600J .

(f)

To determine

The final kinetic energy of the object.

(f)

Expert Solution
Check Mark

Answer to Problem 58P

The final kinetic energy of the object is 674J .

Explanation of Solution

Given information: The mass of the object is 3kg , the velocity of the object is 7j^m/s and the net force acting on the object is 12i^N . The time duration is 5s .

Write the expression to calculate the final kinetic energy of the object.

E=12mvf2=12mvfvf (6)

Substitute 3kg for m and (20i^+7j^)m/s for vf in equation (6) to find E .

E=123kg(20i^+7j^)m/s(20i^+7j^)m/s=123kg(400+49)m/s=673.5J674J

Thus, the final kinetic energy of the object is 674J .

Conclusion:

Therefore, the final kinetic energy of the object is 674J .

(g)

To determine

The final kinetic energy of the object.

(g)

Expert Solution
Check Mark

Answer to Problem 58P

The final kinetic energy of the object is 674J .

Explanation of Solution

Given information: The mass of the object is 3kg , the velocity of the object is 7j^m/s and the net force acting on the object is 12i^N . The time duration is 5s .

Write the expression to calculate the final kinetic energy of the object.

E=12mvi2+W (7)

Substitute 3kg for m , 600J for W and 7j^m/s for vi in equation (7) to find E .

E=123kg(7j^m/s)(7j^m/s)+600J=673.5J674J

Thus, the final kinetic energy of the object is 674J .

Conclusion:

Therefore, the final kinetic energy of the object is 674J .

(h)

To determine

The result of comparison of the answers in part (b), (c) and (f), (g).

(h)

Expert Solution
Check Mark

Answer to Problem 58P

The value of acceleration in part (b), (c) and kinetic energy in part (f), (g) are same.

Explanation of Solution

Given information: The mass of the object is 3kg , the velocity of the object is 7j^m/s and the net force acting on the object is 12i^N . The time duration is 5s .

Write the expression to calculate the acceleration of the object.

a=(vfvi)Δt (2)

Write the expression to calculate the acceleration of the object.

a=Fm (3)

According to the second law of motion,

F=m(vfvi)Δt

Substitute m(vfvi)Δt for F in equation (2).

a=m(vfvi)Δtm=(vfvi)Δt (8)

The equation (2) and (8) are same therefore, the value of acceleration in part (b) and (c) are same.

Write the expression to calculate the work done on the object,

W=changeinkineticenergy=12mvf212mvi2 (9)

Substitute 12mvf212mvi2 for W in equation (9).

E=12mvi2+12mvf212mvi2=12mvf2 (10)

The equation (10) and (6) are same.

Thus, the value of kinetic energy in part (f) and (g) are same.

Conclusion:

Therefore, the value of acceleration in part (b), (c) and kinetic energy in part (f), (g) are same.

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Chapter 8 Solutions

Principles of Physics: A Calculus-Based Text

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Impulse Derivation and Demonstration; Author: Flipping Physics;https://www.youtube.com/watch?v=9rwkTnTOB0s;License: Standard YouTube License, CC-BY