EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
10th Edition
ISBN: 8220106740163
Author: SERWAY
Publisher: CENGAGE L
Question
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Chapter 7, Problem 45AP

(a)

To determine

The expression for the two forces in unit vector notation.

(a)

Expert Solution
Check Mark

Answer to Problem 45AP

The expression for the first force in unit vector notation is (20.5i^+14.3j^)N and the expression for the second force in unit vector notation is (36.4i^+21.0j^)N.

Explanation of Solution

The mass of an object is 5.00kg, the magnitude of the first force is 25.0N at an angle of 35.0° and the magnitude of the second force is 42.0N at an angle of 150°. The velocity of the object at origin is vi=(4.00i^+2.50j^)m/s.

Write the formula to calculate the expression for the first force in unit vector notation is

    F1=(F1cosθ1)i^+(F1sinθ1)j^                                                          (I)

Here, F1 is the unit vector notation of the first force, F1 is the magnitude of the first force and θ1 is the angle made by the first force from the horizontal.

Write the formula to calculate the expression for the second force in unit vector notation

    F2=(F2cosθ2)i^+(F2sinθ2)j^                                                         (II)

Here, F2 is the unit vector notation of the second force, F2 is the magnitude of the second force and θ2 is the angle made by the second force from the horizontal.

Conclusion:

Substitute 25.0N for F1 and 35.0° for θ1 in equation (I) to find F1.

    F1=((25.0N)cos35.0°)i^+((25.0N)sin35.0°)j^=(20.5i^+14.3j^)N

Substitute 42.0N for F2 and 150° for θ2 in equation (II) to find F2

    F2=((42.0N)cos150°)i^+((42.0N)sin150°)j^=(36.4i^+21.0j^)N

Therefore, the expression for the first force in unit vector notation is (20.5i^+14.3j^)N and the expression for the second force in unit vector notation is (36.4i^+21.0j^)N.

(b)

To determine

The total force exerted on the object.

(b)

Expert Solution
Check Mark

Answer to Problem 45AP

The total force exerted on the object is (15.9i^+35.3j^)N.

Explanation of Solution

Write the formula to calculate the total force exerted on the object

    F=F1+F2

Here, F is the total force exerted on the object.

Conclusion:

Substitute (20.5i^+14.3j^)N for F1 and (36.4i^+21.0j^)N for F2.

    F=(20.5i^+14.3j^)N+(36.4i^+21.0j^)N=(15.9i^+35.3j^)N

Therefore, the total force exerted on the object is (15.9i^+35.3j^)N.

(c)

To determine

The acceleration on the object.

(c)

Expert Solution
Check Mark

Answer to Problem 45AP

The acceleration on the object is (3.18i^+7.06j^)N.

Explanation of Solution

Write the formula to calculate the acceleration of the object

    F=ma

Here, a is the acceleration of the object.

Conclusion:

Substitute (15.9i^+35.3j^)N for F and 5.00kg for m to find a.

    (15.9i^+35.3j^)N=(5.00kg)aa=15.00kg(15.9i^+35.3j^)N=(3.18i^+7.06j^)N

Therefore, the acceleration on the object is (3.18i^+7.06j^)N.

(d)

To determine

The velocity on the object.

(d)

Expert Solution
Check Mark

Answer to Problem 45AP

The velocity on the object is (5.54i^+23.7j^)m/s.

Explanation of Solution

Write the formula to calculate the velocity of the object at t=3.00s

    vf=vi+at

Here, vf is the velocity vector of the object at t=3.00s, vi is the velocity vector of the object at origin and t is the time interval during the motion of the object.

Conclusion:

Substitute (4.00i^+2.50j^)m/s for vi, (3.18i^+7.06j^)N for a and 3.00s for t to find vf.

    vf=(4.00i^+2.50j^)m/s+[(3.18i^+7.06j^)N](3.00s)=(4.00i^+2.50j^)m/s+(9.54i^+21.18j^)m/s=(5.54i^+23.7j^)m/s

Therefore, the velocity on the object is (5.54i^+23.7j^)m/s.

(e)

To determine

The position on the object.

(e)

Expert Solution
Check Mark

Answer to Problem 45AP

The position on the object is (2.3i^+39.3j^)m.

Explanation of Solution

Write the formula to calculate the position of the object

    r=vit+12at2

Here, r is the position vector of the object at t=3.00s.

Substitute (4.00i^+2.50j^)m/s for vi, (3.18i^+7.06j^)N for a and 3.00s for t to find r.

    r=[(4.00i^+2.50j^)m/s](3.00s)+12[(3.18i^+7.06j^)N](3.00s)2=(12.0i^+7.50j^)m/s+(14.31i^+31.77j^)m/s=(2.3i^+39.3j^)m/s

Conclusion:

Therefore, the position on the object is (2.3i^+39.3j^)m.

(f)

To determine

The kinetic energy of the object from the formula 12mvf2.

(f)

Expert Solution
Check Mark

Answer to Problem 45AP

The kinetic energy of the object from the formula 12mvf2 is

Explanation of Solution

Write the formula to calculate the magnitude of the final velocity of the object

    vf=vfx2+vfy2

Here, vf is the magnitude of the final velocity of the object.

Substitute 5.54m/s for vfx and 23.7m/s for vfy to find vf.

    vf=(5.54m/s)2+(23.7m/s)2=24.34m/s

Write the formula to calculate the kinetic energy of the object

    Kf=12mvf2

Conclusion:

Substitute 5.00kg for m and 24.34m/s for vf to find Kf.

    Kf=12×5.00kg×(24.34m/s)2=1480.95kgm2/s2×1J1kgm2/s2×103kJ1J=1.48kJ

Therefore, the kinetic energy of the object from the formula 12mvf2 is 1.48kJ.

(g)

To determine

The kinetic energy of the object from the formula 12mvi2+FΔr.

(g)

Expert Solution
Check Mark

Answer to Problem 45AP

The kinetic energy of the object from the formula 12mvi2+FΔr is 1.48kJ.

Explanation of Solution

Write the formula to calculate the magnitude of the initial velocity of the object

    vi=vix2+viy2

Here, vi is the magnitude of the final velocity of the object.

Substitute 4.00m/s for vix and 2.50m/s for viy to find vi.

    vi=(4.00m/s)2+(2.50m/s)2=4.72m/s

Write the formula to calculate the final kinetic energy of the object

    Kf=12mvi2+FΔr

Conclusion:

Substitute 5.00kg for m, 4.72m/s for vf and (15.9i^+35.3j^)N for F and (2.3i^+39.3j^)m for r to find Kf in above expression.

    Kf=12×5.00kg×(4.72m/s)2+[((15.9i^+35.3j^)N)((2.3i^+39.3j^)m)]=55.625kgm2/s2+[36.57+1387.29]Nm×1kgm2/s21Nm=1479.5kgm2/s2×1J1kgm2/s2×103kJ1J=1.48kJ

Therefore, the kinetic energy of the object from the formula 12mvf2 is 1.48kJ.

(h)

To determine

The conclusion by comparing the answer of part (f) and (g).

(h)

Expert Solution
Check Mark

Explanation of Solution

Newton gave the law for the constant acceleration motion while the work energy theorem relates the work done by the object to its energy.

In part (f) the kinetic energy of the object is calculated with the help of Newton’s law while the kinetic energy in part (g) is calculated by the work energy theorem. Since in both the parts the kinetic energy of the object comes out to be same that conclude both the law and theorem are relevant to each other. The work energy theorem is consistent with the Newton’s law.

Conclusion:

Therefore, the work energy theorem is consistent with the Newton’s law of equation.

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Chapter 7 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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