Bundle: Physics for Scientists and Engineers, Volume 2, Loose-leaf Version, 10th + WebAssign Printed Access Card, Single-Term
Bundle: Physics for Scientists and Engineers, Volume 2, Loose-leaf Version, 10th + WebAssign Printed Access Card, Single-Term
10th Edition
ISBN: 9781337888745
Author: SERWAY, Raymond A., Jewett, John W.
Publisher: Cengage Learning
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Chapter 9, Problem 51AP

Review. There are (one can say) three coequal theories of motion for a single particle: Newton’s second law, stating that the total force on the particle causes its acceleration; the work–kinetic energy theorem, stating that the total work on the particle causes its change in kinetic energy; and the impulse–momentum theorem, stating that the total impulse on the panicle causes its change in momentum. In this problem, you compare predictions of the three theories in one particular case. A 3.00-kg object has velocity 7.00 j ^ m / s . Then, a constant net force 12.0 i ^ N acts on the object for 5.00 s. (a) Calculate the object’s final velocity, using the impulse–momentum theorem. (b) Calculate its acceleration from a = ( v f v i ) / Δ t . (c) Calculate its acceleration from a = F / m . (d) Find the object’s vector displacement from Δ r = v i t + 1 2 a t 2 (e) Find the work done on the object from W = F Δ r . (f) Find the final kinetic energy from 1 2 m v f 2 = 1 2 m v f v f . (g) Find the final kinetic energy from 1 2 m v i 2 + W . (h) State the result of comparing the answers to parts (b) and (c), and the answers to parts (f) and (g).

(a)

Expert Solution
Check Mark
To determine

The final velocity of the object.

Answer to Problem 51AP

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

Explanation of Solution

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 and Δt is the time duration.

Conclusion:

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.

(b)

Expert Solution
Check Mark
To determine

The acceleration of the object.

Answer to Problem 51AP

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

Explanation of Solution

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)

Expert Solution
Check Mark
To determine

The acceleration of the object.

Answer to Problem 51AP

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

Explanation of Solution

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)

Expert Solution
Check Mark
To determine

The vector displacement of the object.

Answer to Problem 51AP

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

Explanation of Solution

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)

Expert Solution
Check Mark
To determine

The work done on the object.

Answer to Problem 51AP

The work done on the object is 600J.

Explanation of Solution

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)

Expert Solution
Check Mark
To determine

The final kinetic energy of the object.

Answer to Problem 51AP

The final kinetic energy of the object is 674J.

Explanation of Solution

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)

Expert Solution
Check Mark
To determine

The final kinetic energy of the object.

Answer to Problem 51AP

The final kinetic energy of the object is 674J.

Explanation of Solution

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)

Expert Solution
Check Mark
To determine

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

Answer to Problem 51AP

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

Explanation of Solution

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 9 Solutions

Bundle: Physics for Scientists and Engineers, Volume 2, Loose-leaf Version, 10th + WebAssign Printed Access Card, Single-Term

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