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Chapter 6, Problem 67P

(a)

To determine

To express two vectors in unit vector notation.

(a)

Expert Solution
Check Mark

Answer to Problem 67P

The two vectors in unit vector notation is F1=(20.5i^+14.3j^)N, and F2=(36.4i^+21.0j^)N.

Explanation of Solution

Write the expression for force from Figure P6.67.

    F1=(25.0N)(cos35.0°i^+sin35.0°j^)

Simplify the above equation.

    F1=(20.5i^+14.3j^)N        (I)

Write the expression for force from Figure P6.67.

    F2=(42.0N)(cos150°i^+sin150°j^)

Simplify the above equation.

    F2=(36.4i^+21.0j^)N        (II)

Conclusion:

Therefore, the two vectors in unit vector notation is F1=(20.5i^+14.3j^)N, and F2=(36.4i^+21.0j^)N.

(b)

To determine

The total force exerted on the object.

(b)

Expert Solution
Check Mark

Answer to Problem 67P

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

Explanation of Solution

Write the total force exerted on the object.

    F=F1+F2        (III)

Use equation (I) and (II) in equation (III), to find the net force.

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

Conclusion:

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

(c)

To determine

The acceleration of the object.

(c)

Expert Solution
Check Mark

Answer to Problem 67P

The acceleration of the object is (3.18i^+7.07j^)m/s2_.

Explanation of Solution

Write the expression for acceleration of the object.

    a=Fm        (V)

Here, a is the acceleration, m is the mass, and F is the net force on the object.

Use equation (IV) in (V).

    a=(15.9i^+35.3j^)Nm        (VI)

Given that the mass of the object is 5kg. Apply this condition in equation (VI).

    a=(15.9i^+35.3j^)N5kg

Rearrange the above equation.

    a=(3.18i^+7.07j^)m/s2        (VII)

Conclusion:

Therefore, the acceleration of the object is (3.18i^+7.07j^)m/s2_.

(d)

To determine

The velocity of the object.

(d)

Expert Solution
Check Mark

Answer to Problem 67P

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

Explanation of Solution

Write the velocity of the object.

    vf=vi+at        (VIII)

Here, vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.

Conclusion:

Substitute (4.00i^+2.50j^)m/s for vi, (3.18i^+7.07j^)m/s2 for a, and 3.00s for t in equation (VIII), to find vf.

    vf=(4.00i^+2.50j^)m/s+(3.18i^+7.07j^)m/s2×3.00s=(5.54i^+23.7j^)m/s

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

(e)

To determine

The position of the object at time 3.00s.

(e)

Expert Solution
Check Mark

Answer to Problem 67P

The position of the object at 3.00s is (2.30i^+39.3j^)m_.

Explanation of Solution

Given that the initial position of the object is zero.

Write the expression for position using equation of motion.

    rf=ri+vit+12at2        (IX)

Here, rf is the final position, ri is the initial position, vi is the initial velocity, t is the time, and, a is the acceleration.

Conclusion:

Substitute (4.00i^+2.50j^)m/s for vi, (3.18i^+7.07j^)m/s2 for a, 3.00s for t, 0m for ri, in equation (IX), to find rf.

    rf=(0m)+(4.00i^+2.50j^)m/s(3.00s)+12(3.18i^+7.07j^)m/s2×(3.00s)2=(2.30i^+39.3j^)m

Therefore, the position of the object at 3.00s is (2.30i^+39.3j^)m_.

(f)

To determine

The final kinetic energy of the object.

(f)

Expert Solution
Check Mark

Answer to Problem 67P

The final kinetic energy of the object is 1.48kJ_.

Explanation of Solution

Write the expression for kinetic energy.

    Kf=12mvf2        (X)

Here, Kf is the final kinetic energy, m is the mass, and vf is the final velocity.

From subpart (d) the final velocity vector is given by

  vf=(5.54i^+23.7j^)m/s

Write the magnitude of final velocity vector.

    |vf|=(5.54)2+(23.7)2m/s        (XI)

Conclusion:

Substitute 5.00kg for m, and (5.54)2+(23.7)2m/s for vf in equation (X), to find Kf.

    Kf=12(5.00kg)((5.54)2+(23.7)2m/s)2=1480.95J=1480.95J×1kJ103J=1.48kJ

Therefore, the final kinetic energy of the object is 1.48kJ_.

(g)

To determine

The final kinetic energy of the object using the equation Kf=12mvf2+F.Δr.

(g)

Expert Solution
Check Mark

Answer to Problem 67P

The final kinetic energy of the object using the equation Kf=12mvf2+F.Δr is 1.48kJ_.

Explanation of Solution

Write the expression for kinetic energy.

  Kf=12mvf2+F.Δr        (XII)

From subpart (b) the net force F=(15.9i^+35.3j^)N, and from subpart (e) the change in position is Δr=(2.30i^+39.3j^)m.

Apply the above condition in equation (XII).

  Kf=12mvf2+[(15.9i^+35.3j^)N(2.30i^+39.3j^)m]        (XIII)

Conclusion:

Substitute 5.00kg for m, and (5.54)2+(23.7)2m/s for vf in equation (XIII), to find Kf.

    Kf=12(5.00kg)((5.54)2+(23.7)2m/s)2+[(15.9i^+35.3j^)N(2.30i^+39.3j^)m]=55.6J+1426J=1480J×1kJ103J=1.48kJ

Therefore, the final kinetic energy of the object using the equation Kf=12mvf2+F.Δr is 1.48kJ_.

(h)

To determine

To compare the answers in subpart (f) and subpart (g).

(h)

Expert Solution
Check Mark

Answer to Problem 67P

The answers in subpart (f) and subpart (g) is same, and it proves that the work energy theorem is consistent with the newton’s second law of motion.

Explanation of Solution

From subpart (f) and (g), the kinetic energy is obtained as 1.48kJ.

The work energy theorem states that the change in kinetic energy is equal to the external work done, and by rearranging the terms the final kinetic energy of the object is obtained in subpart (g).

Since the answers in subpart (g) and (f) are same, which proves that the work energy theorem is consistent with the newton’s second law of motion.

Conclusion:

Therefore, the answers in subpart (f) and subpart (g) is same, and it proves that the work energy theorem is consistent with the newton’s second law of motion.

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Chapter 6 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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