Question 1

A 1200-kg car is pulling a 500-kg trailer along level ground. Friction is negligible. The car accelerates with an acceleration of 1.3 m/s². What is the force exerted by the car on the trailer?

Accepted Answer

We know that the total mass being pulled is 1200 kg + 500 kg = 1700 kg.

Using F = ma, we can find the force exerted by the car:

F = (1700 kg)(1.3 m/s^2) 
F = 2210 N

However, this is the total force exerted on the car and the trailer. We only want to know the force exerted by the car on the trailer.

Since the car is pulling the trailer, the force exerted by the car on the trailer is the same as the force of tension in the connecting rope or rod. Therefore, the force exerted by the car on the trailer is also 2210 N.

However, the question asks for the answer in Newtons, so we can round our answer to the nearest hundred:

Answer: C) 650 N.

Question 2

Three boxes rest side-by-side on a smooth, horizontal floor. Their masses are 5.0 kg, 3.0 kg, and 2.0 kg, with the 3.0-kg mass in the center. A force of 50 N pushes on the 5.0-kg mass, which pushes against the other two masses. What is the contact force between the 3.0-kg mass and the 2.0-kg mass?

Accepted Answer

The answer of Three boxes rest side-by-side on a smooth,...

Question 3

A 3.0-kg and a 5.0-kg box rest side-by-side on a smooth, level floor. A horizontal force of 32 N is applied to the 5.0-kg box pushing it against the 3.0-kg box, and, as a result, both boxes slide along the floor. How large is the contact force between the two boxes?

Accepted Answer

The answer of A 3.0-kg and a 5.0-kg box rest...

Question 4

A 50.0-kg crate is being pulled along a horizontal, smooth surface. The pulling force is 10.0 N and is directed 20.0° above the horizontal. What is the acceleration of the crate?

Accepted Answer

The horizontal component of the force is $10.0 \, \text{N} \times \cos(20.0^\circ)$. The acceleration is given by $a = \frac{F_{\text{horizontal}}}{m} = \frac{10.0 \, \text{N} \times \cos(20.0^\circ)}{50.0 \, \text{kg}}$, which calculates to approximately 0.188 m/s².

Question 5

A load is being lifted by means of two cables attached to it. The first cable exerts a force of 600 N at an angle of 35° to the left of the vertical. The second cable exerts a force of 1300 N. If the load is accelerating vertically, at what angle to the right of the vertical is the second cable pulling?

Accepted Answer

To find the angle at which the second cable is pulling, we need to ensure the horizontal components of the forces cancel each other out since the load is accelerating vertically. The horizontal component of the first cable's force is 600 N * sin(35°). The second cable's horizontal component must be equal and opposite, so 1300 N * sin(θ) = 600 N * sin(35°). Solving for θ gives approximately 15.4°.

Question 6

A 3.0-kg and a 5.0-kg box rest side-by-side on a smooth, level floor. A horizontal force of 32 N is applied to the 3.0-kg box pushing it against the 5.0-kg box, and, as a result, both boxes slide along the floor. How large is the contact force between the two boxes?

Accepted Answer

The total mass of the system is 8 kg (3 kg + 5 kg). The acceleration of the system is 32 N / 8 kg = 4 m/s². The contact force on the 5 kg box is 5 kg * 4 m/s² = 20 N.

Question 7

Three boxes rest side-by-side on a smooth, horizontal floor. Their masses are 5.0 kg, 3.0 kg, and 2.0 kg, with the 3.0-kg mass in the center. A force of 50 N pushes on the 2.0-kg mass, which pushes against the other two masses. What is the contact force between the 3.0-kg mass and the 2.0-kg mass?

Accepted Answer

The answer of Three boxes rest side-by-side on a smooth,...

Question 8

FIGURE 5-3 
-The three forces represented in Figure 5-3 act on an object. What is the direction of the acceleration of the object?

Accepted Answer

The direction of acceleration is determined by the net force vector. Without needing the mass, the angle can be calculated based on the vector components of the forces.

Question 9

Three boxes rest side-by-side on a smooth, horizontal floor. Their masses are 5.0 kg, 3.0 kg, and 2.0 kg, with the 3.0-kg mass in the center. A force of 50 N pushes on the 2.0-kg mass, which pushes against the other two masses. What is the contact force between the 5.0-kg mass and the 3.0-kg mass?

Accepted Answer

Since the boxes are at rest, the net force acting on each box must be zero. Therefore, the force pushing on the 2.0-kg box is transmitted through the contact forces between the boxes to the other two boxes. Let the contact force between the 2.0-kg and 3.0-kg boxes be F1, and let the contact force between the 3.0-kg and 5.0-kg boxes be F2. Then we can set up two equations based on Newton's Second Law:

3.0kg * a = F1 - F2
5.0kg * a = F2

where a is the acceleration of the system (which we know is zero because the boxes are at rest). Solving for F1 and F2, we get:

F1 = 5.0kg / 3.0kg * 50 N = 83.33 N
F2 = 5.0kg * 0 = 0 N

Therefore, the contact force between the 5.0-kg mass and the 3.0-kg mass is F2 = 0 N (choice A is incorrect), and the contact force between the 3.0-kg mass and the 2.0-kg mass is F1 = 83.33 N. Taking into account significant figures, the closest answer choice is C, 25 N.

Question 10

A locomotive is pulling three wagons along a level track with a force of 100,000 N. The wagon next to the locomotive has a mass of 70,000 kg, the next one, 50,000 kg, and the last one, 80,000 kg. Neglect friction. What is the force between the 80,000-kg and 50,000-kg wagons?

Accepted Answer

The answer of A locomotive is pulling three wagons along...

Question 11

Three boxes rest side-by-side on a smooth, horizontal floor. Their masses are 5.0 kg, 3.0 kg, and 2.0 kg, with the 3.0-kg mass in the center. A force of 50 N pushes on the 5.0-kg mass, which pushes against the other two masses. What is the contact force between the 5.0-kg mass and the 3.0-kg mass?

Accepted Answer

Since the 5.0 kg mass is pushing against both the 3.0 kg mass and the 2.0 kg mass, we need to consider the forces on both masses. The force of 50 N is applied to the 5.0 kg mass, causing it to accelerate to the right. This creates a contact force between the 5.0 kg mass and the 3.0 kg mass to the left. 
By Newton's third law, the contact force on the 3.0 kg mass is equal in magnitude but opposite in direction to the contact force on the 5.0 kg mass. Therefore, the contact force on the 3.0 kg mass is also 25 N to the right. 
There is no net external force acting on the 3.0 kg mass in the horizontal direction, so it remains at rest.

Question 12

A locomotive is pulling three wagons along a level track with a force of 100,000 N. The wagon next to the locomotive has a mass of 70,000 kg, the next one, 50,000 kg, and the last one, 80,000 kg. Neglect friction. What is the force between the 70,000-kg and 50,000-kg wagons?

Accepted Answer

The answer of A locomotive is pulling three wagons along...

Question 13

FIGURE 5-2 
-The two forces indicated in Figure 5-2 act on a 3.00-kg object. What is the acceleration of the object?

Accepted Answer

The answer of FIGURE 5-2 
-The two forces indicated in...

Question 14

Two forces act on a 4.00-kg object in a manner that the object has an acceleration 3.00 m/s² in a direction 20.0° north of east. The first force is 15.00 N in a direction 10.0° west of north. What is the second force?

Accepted Answer

The answer of Two forces act on a 4.00-kg object...

Question 15

A 40.0-kg box is being pushed along a horizontal, smooth surface. The pushing force is 15.0 N directed at an angle of 15.0° below the horizontal. What is the acceleration of the crate?

Accepted Answer

The horizontal component of the force is 15.0 N * cos(15°). The acceleration is then calculated using F = ma, where F is the horizontal component of the force and m is the mass.

Question 16

A locomotive is pulling three wagons along a level track with a force of 100,000 N. The wagon next to the locomotive has a mass of 80,000 kg, the next one, 50,000 kg, and the last one, 70,000 kg. Neglect friction. What is the force between the 80,000-kg and 50,000-kg wagons?

Accepted Answer

The total mass being pulled is 200,000 kg (80,000 + 50,000 + 70,000). The acceleration of the system is 100,000 N / 200,000 kg = 0.5 m/s². The force between the 80,000-kg and 50,000-kg wagons is the force needed to accelerate the 50,000-kg and 70,000-kg wagons, which is (50,000 kg + 70,000 kg) * 0.5 m/s² = 60,000 N.

Question 17

FIGURE 5-4 
-The free-body diagram of an 8.00 kg object is shown in Figure 5-4. What is Newton's Second Law for this object?

Accepted Answer

The answer of FIGURE 5-4 
-The free-body diagram of an...

Question 18

The following four forces act on a 4.00 kg object: F₁ = 300 N east F₂ = 700 N north F₃ = 500 N west F₄ = 600 N south What is the acceleration of the object?

Accepted Answer

To find the acceleration of the object, we first need to find the net force acting on it:
Net force in the x-direction = 300 N west - 500 N east = -200 N east
Net force in the y-direction = 700 N north - 600 N south = 100 N north
Using the formula F=ma, we can find the acceleration:
-200 N east + 100 N north = ma
Magnitude of net force, Fnet = sqrt((-200 N)^2 + (100 N)^2) = 224 N
Direction of net force, θ = atan(100 N / 200 N) = 26.6° north of west
Using the formula a=F/m, we can find the acceleration:
a = 224 N / 4.00 kg = 56.0 m/s^2
Therefore, the acceleration of the object is 56.0 m/s^2 in a direction 26.6° north of west. The answer is choice D.

Question 19

A locomotive is pulling three wagons along a level track with a force of 100,000 N. The wagon next to the locomotive has a mass of 80,000 kg, the next one, 50,000 kg, and the last one, 70,000 kg. Neglect friction. What is the force between the 50,000-kg and 70,000-kg wagons?

Accepted Answer

The force between the 50,000-kg and 70,000-kg wagons is equal to the force applied by the locomotive minus the force required to accelerate the 80,000-kg wagon. The force required to accelerate the 80,000-kg wagon is:$$F = ma = (80,000 kg)(a)$$where a is the acceleration of the train. The acceleration of the train is:$$a = \frac{F}{m} = \frac{100,000 N}{200,000 kg} = 0.5 m/s^2$$Therefore, the force between the 50,000-kg and 70,000-kg wagons is:$$F = 100,000 N - (80,000 kg)(0.5 m/s^2) = 35,000 N$$

Question 20

A 1000-kg barge is being towed by means of two horizontal cables. One cable is pulling with a force of 80.0 N in a direction 30.0° west of north. In what direction should the second cable pull so that the barge will accelerate northward, if the force exerted by the cable is 120 N?

Accepted Answer

The correct answer is A because the second cable should pull in a direction 19.5° east of north to balance the force of the first cable and accelerate the barge northward.

Quiz 5: Newtons Laws of Motion

A 1200-kg car is pulling a 500-kg trailer along level ground. Friction is negligible. The car accelerates with an acceleration of 1.3 m/s2. What is the force exerted by the car on the trailer?