Section 3.3Newton's Second Law of Motion

1. (Exercises 1 and 2) A man experiences an acceleration of 0.250 m/s² when his feet push backwards against the sidewalk with a horizontal force of 42.0 N. What is the man's mass?

F = ma;   m = F / a = 42.0 N / 0.250 m/s² = 168 kg

2. (Exercises 3 and 4) What will be the acceleration of a 6.00-kg package of nails if it has a force of 58.8 N downward acting on it as it falls towards the ground?

a = F / m = 58.8 N / 6.00 kg = 9.80 m/s² (downward)

Note: This result should have been expected because the acceleration of any freely falling object near Earth's surface will be equal to g, which we already know is equal to 9.80 m/s².

3. (Exercises 7 and 8) What is the weight in newtons of a 5.0-kg fish?

Use the equation w = mg. The mass is 5.0 kg, and g is 9.80 m/s². Thus the weight is:

w = (5.0 kg) (9.80 m/s²) = 49 kg-m/s² = 49 N

The weight of the fish is 49 newtons.

Section 3.5Newton's Law of Gravitation

4. (Exercises 9 and 10) Two students are standing 2.00 m apart on Earth's surface. One student has a mass of 45.0 kg and the other has a mass of 60.0 kg.

a. How much gravitational force is exerted on the 45.0-kg student by the 60.0-kg student?

F = G m₁ m₂ / r ²
= (6.67 x 10^-11-m²/kg²)(45.0 kg)(60.0 kg) / (2.00 m)²
= 4.50 x 10^-8 N

b. How much force is exerted on the 60.0-kg student by the 45.0-kg student?

Here we could perform the calculation again, reversing the values for m₁ and m₂ in the above equation, but the results would be the same: 4.50 x 10^-8 N. It is also possible to use Newton's third law here. Because the two students exert equal and opposite forces on each other, as soon as one of the forces is calculated, the other is immediately known. Remember that the force of the second student is the same magnitude, but is in the opposite direction. If student 1 were standing east of student 2, the force exerted on student 1 is toward the west, while the force exerted on student 2 is toward the east.

c. How much force does Earth exert on each student as they stand there?

We could again use Newton's law of gravitation to solve this problem, where

m₂ = M_E and r is the Earth's radius, but it is much simpler to use the acceleration due to gravity at Earth's surface, which we know to be 9.80 m/s², to find the weight of each student.

F₁ = w = m₁ g = 45.0 kg (9.80 m/s²) = 441 N and

F₂ = w = m₂ g = 60.0 kg (9.80 m/s²) = 588 N

Note that the weight of each student is different because they have different masses. These two forces are also not in opposite directions; in this case, they are both toward the center of Earth, or downward.

5. (Exercises 13 and 14) If a car and its passengers having a total mass of 1386 kg were on the Moon, what would be the total weight, in newtons, of the car and passengers?

w = mg, but g is not 9.80 m/s² on the Moon; it is only 1/6 that value, so:

g_Moon = 1/6 (9.80 m/s²) = 1.63 m/s²

w_Moon = mg_Moon = 1386 kg (1.63 m/s²) = 2260 N

The total weight of the car and its passengers is 2260 newtons.

Section 3.6Momentum

6. (Exercises 15 and 16) Determine the linear momentum of a runner who has a weight on Earth of 735 N, when he is running at a rate of 1.20 m/s toward the west.

p = mv = (w/g) v = (735 N / 9.80 m/s²) 1.20 m/s = 90.0 N-s, to the west

Note that either the units "N-s" or "kg-m/s" can be used for linear momentum and that, because momentum is a vector quantity, its direction must be specified; in this case "to the west."

7. (Exercises 19 and 20) The Moon travels around Earth on a path that is not a perfect circle but an ellipse. The Moon's closest approach to Earth (perigee) is 3.57 x 10⁸ m, at which time if has a speed of 1080 m/s. At what speed does it travel in its orbit when it is at apogee (the farthest point in its elliptical orbit), if this distance is 4.08 x 10⁸ m?

Conservation of angular momentum is needed to solve this problem.

m v₁ r₁ = m v₂ r₂

Solving for v₂ we find:

v₂ = m v₁ r₁ / m r₂

Because m is the same, it cancels out and the result will not depend on the value of the mass.

v₂ = v₁ r₁ / r₂ = (1080 m/s) (3.57 x 10⁸ m) / 4.08 x 10⁸ m = 945 m/s

The conservation of angular momentum requires that an orbiting object like the Moon travel faster when it is farther from the center of its orbit than it does when it is closest. This problem clearly shows this concept.

Notice that the 10⁸ is in both the numerator and denominator of the final equation and so cancels out nicely, and the math becomes quite simple. This is a handy tool to remember if the powers-of-10 values are the same in the numerator and denominator in other exercises, as they are here.

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