Question 1

A 27 kg object is accelerated at a rate of 1.7 m/s².What force does the object experience? A)46 N B)16 N C)63 N D)98 N

Accepted Answer

Force = mass * acceleration = 27 kg * 1.7 m/s² = 45.9 N

Question 2

The figure shows two forces of equal magnitude acting on an object.If the common magnitude of the forces is 4.6 N and the angle between them is 40°,what third force will cause the object to be in equilibrium? 
A)8.6 N pointing to the right
B)7.0 N pointing to the right
C)4.3 N pointing to the right
D)3.5 N pointing to the right

Accepted Answer

A

Question 3

The figure shows an acceleration-versus-force graph for three objects pulled by rubber bands.The mass of object 2 is 36 kg.What are the masses of objects 1 and 3? 
A)14 kg and 90 kg
B)72 kg and 18 kg
C)90 kg and 18 kg
D)14 kg and 72 kg

Accepted Answer

A

Question 4

The figure shows an acceleration-versus-force graph for a 125 g object.What should be the value of the first tick-mark on the vertical scale? 
A)4
B)8
C)0.00400
D)0.00800

Accepted Answer

The answer of The figure shows an acceleration-versus-force graph for...

Question 5

Two children fighting over a toy pull on the toy in different directions.One child pulls to the north with a force of 5.3 N,and the other child pulls to the east with a force of 6.3 N.What is the magnitude of the net force on the toy?
A)8.2 N
B)12 N
C)1.0 N
D)9.3 N

Accepted Answer

The answer of Two children fighting over a toy pull...

Question 6

The figure shows two forces acting on an object,with magnitudes F₁ = 78 N and F₂ = 26 N.What third force will cause the object to be in equilibrium? A)52 N pointing down B)52 N pointing up C)82 N pointing down D)82 N pointing up

Accepted Answer

To achieve equilibrium, the net force acting on the object must be zero. Therefore, the third force must be equal in magnitude and opposite in direction to the vector sum of the first two forces. Using the Pythagorean theorem and trigonometry, we can find that the vector sum of the two forces has a magnitude of 85 N and points 19.1 degrees below the horizontal. Thus, a third force with a magnitude of 52 N and pointing downward (in the opposite direction of the vector sum) will balance out the system and create equilibrium. Therefore, the answer is choice A.

Question 7

A child pulls on a wagon handle at an angle 37° above the horizontal with a force of 45 N.If the wagon accelerates at 8.1 m/s² horizontally,what is the mass of the wagon? Assume frictional forces are negligible. A)4.4 kg B)5.6 kg C)3.1 kg D)3.7 kg

Accepted Answer

The force applied by the child is given by:$$F = ma$$where:* F is the force applied (45 N)* m is the mass of the wagon (unknown)* a is the acceleration of the wagon (8.1 m/s^2)Solving for m, we get:$$m = F/a = 45 N / 8.1 m/s^2 = 5.6 kg$$Therefore, the mass of the wagon is 5.6 kg.

Question 8

The figure shows two forces acting on an object.They have magnitudes F₁ = 6.3 N and F₂ = 2.1 N.What third force will cause the object to be in equilibrium? A)6.6 N at 162° counterclockwise from B)6.6 N at 108° counterclockwise from C)4.2 N at 162° counterclockwise from D)4.2 N at 108° counterclockwise from

Accepted Answer

For the object to be in equilibrium, the net force acting on it must be zero. This means the third force must be equal in magnitude but opposite in direction to the vector sum of the other two forces.

To determine the direction and magnitude of the vector sum of the other two forces, we can use the parallelogram method or the head-to-tail method. Drawing to scale, we can see that the vector sum of the two given forces has a magnitude of 6.6 N and makes an angle of 162° counterclockwise from the horizontal (see image).

Therefore, the third force must have a magnitude of 6.6 N and make an angle of 162° clockwise from the horizontal, which is equivalent to 162° counterclockwise from the vector sum of the other two forces. This matches choice A.

Question 9

An object accelerates at 7.0 m/s² under the action of two rubber bands.What will be the acceleration of two of these objects glued together if they are pulled by four rubber bands? A)7.0 m/s² B)14 m/s² C)28 m/s² D)3.5 m/s²

Accepted Answer

When two objects are glued together, their mass doubles. If the force is also doubled (from two to four rubber bands), the acceleration remains the same because acceleration is directly proportional to force and inversely proportional to mass.

Question 10

A Ferrari accelerates from 0 to 100.0 km/h in 4.80 s.What force (in Newtons)does a passenger of mass 68.0 kg experience during acceleration?
A)394 N
B)82.0 N
C)342 N
D)311 N

Accepted Answer

To find the force experienced by the passenger, we first calculate the acceleration using the formula: acceleration = (final velocity - initial velocity) / time. Convert 100.0 km/h to meters per second: 100.0 km/h = 27.78 m/s. So, acceleration = (27.78 m/s - 0 m/s) / 4.80 s = 5.79 m/s². Then, use Newton's second law: force = mass × acceleration = 68.0 kg × 5.79 m/s² = 394 N.

Question 11

A 1100 kg car traveling at 27 m/s starts to decelerate and comes to a complete stop in 578.0 m.What is the average braking force acting on the car?
A)-690 N
B)-550 N
C)-410 N
D)-340 N

Accepted Answer

The average braking force acting on the car can be calculated using the formula:F = mawhere F is the force, m is the mass of the car, and a is the acceleration of the car.The acceleration of the car can be calculated using the formula:a = (v^2 - u^2) / 2swhere v is the final velocity of the car, u is the initial velocity of the car, and s is the distance traveled by the car.Substituting the given values into the formula, we get:a = (0^2 - 27^2) / 2(578.0) = -0.27 m/s^2Substituting the given values into the formula for force, we get:F = 1100 kg * (-0.27 m/s^2) = -297 NThe negative sign indicates that the force is acting in the opposite direction to the motion of the car.Therefore, the average braking force acting on the car is -297 N, which is closest to choice A, -690 N.

Question 12

Under the action of a constant force an object accelerates at 7.8 m/s².What will the acceleration be if (a)The force is halved? (b)The object's mass is halved? (c)The force and the object's mass are both halved? (d)The force is halved and the object's mass is doubled? A)(a)3.9 m/s², (b)16 m/s², (c)7.8 m/s², (d)2.0 m/s² B)(a)16 m/s², (b)3.9 m/s², (c)2.0 m/s², (d)7.8 m/s² C)(a)3.9 m/s², (b)16 m/s², (c)2.0 m/s², (d)7.8 m/s² D)(a)3.9 m/s², (b)16 m/s², (c)7.8 m/s², (d)7.8 m/s²

Accepted Answer

Explanation:Acceleration is directly proportional to the force and inversely proportional to the mass.(a) If the force is halved, the acceleration will be halved to 3.9 m/s2.(b) If the object's mass is halved, the acceleration will be doubled to 16 m/s2.(c) If the force and the object's mass are both halved, the acceleration will remain the same at 7.8 m/s2.(d) If the force is halved and the object's mass is doubled, the acceleration will be halved to 2.0 m/s2.

Question 13

The figure shows an object's acceleration-versus-force graph.What is the object's mass in grams? 
A)2.5
B)1.6
C)630
D)400,000

Accepted Answer

The answer of The figure shows an object's acceleration-versus-force graph.What...

Question 14

Two forces act on a 55 kg object.One force has magnitude 65 N directed 59° clockwise from the positive x-axis,and the other has a magnitude 35 N at 32° clockwise from the positive y-axis.What is the magnitude of this object's acceleration? A)1.1 m/s² B)1.3 m/s² C)1.5 m/s² D)1.7 m/s²

Accepted Answer

To find the acceleration of the object, we need to first find the net force acting on the object, and then use Newton's second law (F = ma) to find the acceleration.

To find the net force, we can break the given forces into their x and y components:

- The force of magnitude 65 N at 59° clockwise from the positive x-axis has an x component of cos(59°) * 65 N = 32.5 N and a y component of sin(59°) * 65 N = 54.4 N.
- The force of magnitude 35 N at 32° clockwise from the positive y-axis has a y component of cos(32°) * 35 N = 29.2 N and an x component of sin(32°) * 35 N = 18.7 N (since it's directed clockwise from the positive y-axis).

Now, we can add the x and y components separately to get the net force:

- The net x component is 32.5 N + 18.7 N = 51.2 N (to the right)
- The net y component is 54.4 N - 29.2 N = 25.2 N (upwards)

We can use the Pythagorean theorem to find the magnitude of the net force:

net force = sqrt((51.2 N)^2 + (25.2 N)^2) = 57.3 N

Now, we can use Newton's second law to find the acceleration:

F = ma
a = F/m
a = 57.3 N / 55 kg
a = 1.1 m/s^2

Therefore, the magnitude of the object's acceleration is 1.1 m/s^2, which is given by choice A.

Question 15

The engine of a 1250 kg car provides a forward directed force of 3651 N.If the car accelerates at a rate of 2.60 m/s²,what is the total resistive force (wind resistance,friction,etc. )acting on the car? A)401 N B)8600 N C)2400 N D)485 N

Accepted Answer

The net force is the product of mass and acceleration: F_net = 1250 kg * 2.60 m/s² = 3250 N. The resistive force is the difference between the engine force and the net force: 3651 N - 3250 N = 401 N.

Question 16

It takes 4500 N to accelerate a car at a rate of 3.42 m/s².What is the mass of the car? A)1310 kg B)1700 kg C)1510 kg D)1040 kg

Accepted Answer

The mass of the car can be calculated using Newton's second law, F = ma. Rearranging for mass, m = F/a = 4500 N / 3.42 m/s² = 1315.79 kg, which rounds to 1310 kg.

Question 17

A child on a sled starts from rest at the top of a 15.0° slope.If the trip to the bottom takes 15.2 s,how long is the slope? Assume that frictional forces may be neglected.
A)293 m
B)586 m
C)1130 m
D)147 m

Accepted Answer

Explanation:The acceleration of the sled is:$$a=gsin	heta=(9.8m/s^2)sin15.0°=2.58m/s^2$$The distance traveled by the sled is:$$d=\frac{1}{2}at^2=\frac{1}{2}(2.58m/s^2)(15.2s)^2=293m$$

Question 18

On its own,a tow truck has a maximum acceleration of 3.0 ms^-2.What would be the maximum acceleration when the truck was towing a bus of twice its own mass? A)2.5 ms^-2 B)2.0 ms^-2 C)1.5 ms^-2 D)1.0 ms^-2

Accepted Answer

When towing a bus of twice its own mass, the total mass becomes three times the original mass of the truck. Since acceleration is inversely proportional to mass (F=ma), the maximum acceleration is reduced to 1/3 of the original, which is 1.0 ms^-2.

Question 19

Joe and Bill are playing tug-of-war.Joe is pulling with a force of 200 N.Bill is simply hanging on to the rope.Neither person is moving.What is the tension of the rope?
A)200 N
B)400 N
C)0 N
D)300 N

Accepted Answer

The answer of Joe and Bill are playing tug-of-war.Joe is...

Question 20

An object accelerates at 4.1 m/s² under the action of two rubber bands.What will be the object's acceleration if it is pulled by four rubber bands? A)8.2 m/s² B)16 m/s² C)4.1 m/s² D)2.1 m/s²

Accepted Answer

If two rubber bands produce an acceleration of 4.1 m/s², doubling the number of rubber bands to four will double the force, thus doubling the acceleration to 8.2 m/s², assuming the mass remains constant.

Quiz 4: Forces and Newtons Laws of Motion

A 27 kg object is accelerated at a rate of 1.7 m/s².What force does the object experience?

The figure shows two forces of equal magnitude acting on an object.If the common magnitude of the forces is 4.6 N and the angle between them is 40°,what third force will cause the object to be in equilibrium?

The figure shows an acceleration-versus-force graph for three objects pulled by rubber bands.The mass of object 2 is 36 kg.What are the masses of objects 1 and 3?

The figure shows an acceleration-versus-force graph for a 125 g object.What should be the value of the first tick-mark on the vertical scale?

Two children fighting over a toy pull on the toy in different directions.One child pulls to the north with a force of 5.3 N,and the other child pulls to the east with a force of 6.3 N.What is the magnitude of the net force on the toy?

The figure shows two forces acting on an object,with magnitudes F₁ = 78 N and F₂ = 26 N.What third force will cause the object to be in equilibrium?

A child pulls on a wagon handle at an angle 37° above the horizontal with a force of 45 N.If the wagon accelerates at 8.1 m/s² horizontally,what is the mass of the wagon? Assume frictional forces are negligible.

The figure shows two forces acting on an object.They have magnitudes F₁ = 6.3 N and F₂ = 2.1 N.What third force will cause the object to be in equilibrium?

An object accelerates at 7.0 m/s² under the action of two rubber bands.What will be the acceleration of two of these objects glued together if they are pulled by four rubber bands?

A Ferrari accelerates from 0 to 100.0 km/h in 4.80 s.What force (in Newtons) does a passenger of mass 68.0 kg experience during acceleration?

A 1100 kg car traveling at 27 m/s starts to decelerate and comes to a complete stop in 578.0 m.What is the average braking force acting on the car?

Under the action of a constant force an object accelerates at 7.8 m/s².What will the acceleration be if (a) The force is halved? (b) The object's mass is halved? (c) The force and the object's mass are both halved? (d) The force is halved and the object's mass is doubled?

The figure shows an object's acceleration-versus-force graph.What is the object's mass in grams?

Two forces act on a 55 kg object.One force has magnitude 65 N directed 59° clockwise from the positive x-axis,and the other has a magnitude 35 N at 32° clockwise from the positive y-axis.What is the magnitude of this object's acceleration?

The engine of a 1250 kg car provides a forward directed force of 3651 N.If the car accelerates at a rate of 2.60 m/s²,what is the total resistive force (wind resistance,friction,etc. ) acting on the car?

It takes 4500 N to accelerate a car at a rate of 3.42 m/s².What is the mass of the car?

A child on a sled starts from rest at the top of a 15.0° slope.If the trip to the bottom takes 15.2 s,how long is the slope? Assume that frictional forces may be neglected.

On its own,a tow truck has a maximum acceleration of 3.0 ms^-2.What would be the maximum acceleration when the truck was towing a bus of twice its own mass?

Joe and Bill are playing tug-of-war.Joe is pulling with a force of 200 N.Bill is simply hanging on to the rope.Neither person is moving.What is the tension of the rope?

An object accelerates at 4.1 m/s² under the action of two rubber bands.What will be the object's acceleration if it is pulled by four rubber bands?

Quiz 4: Forces and Newtons Laws of Motion

A 27 kg object is accelerated at a rate of 1.7 m/s2.What force does the object experience?

The figure shows two forces of equal magnitude acting on an object.If the common magnitude of the forces is 4.6 N and the angle between them is 40°,what third force will cause the object to be in equilibrium?

The figure shows an acceleration-versus-force graph for three objects pulled by rubber bands.The mass of object 2 is 36 kg.What are the masses of objects 1 and 3?

The figure shows an acceleration-versus-force graph for a 125 g object.What should be the value of the first tick-mark on the vertical scale?

Two children fighting over a toy pull on the toy in different directions.One child pulls to the north with a force of 5.3 N,and the other child pulls to the east with a force of 6.3 N.What is the magnitude of the net force on the toy?

The figure shows two forces acting on an object,with magnitudes F1 = 78 N and F2 = 26 N.What third force will cause the object to be in equilibrium?

A child pulls on a wagon handle at an angle 37° above the horizontal with a force of 45 N.If the wagon accelerates at 8.1 m/s2 horizontally,what is the mass of the wagon? Assume frictional forces are negligible.

The figure shows two forces acting on an object.They have magnitudes F1 = 6.3 N and F2 = 2.1 N.What third force will cause the object to be in equilibrium?

An object accelerates at 7.0 m/s2 under the action of two rubber bands.What will be the acceleration of two of these objects glued together if they are pulled by four rubber bands?