# Practice Test: Forces And Motion

By Junho Song
Junho Song, Physics/Science Teacher
Junho Song, a dedicated educator, specializes in teaching physics and science. Passionate about inspiring students in the fascinating world of scientific discovery and understanding.
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Questions: 20 | Attempts: 4,066  Settings  Sample test preparation for Forces and Motion unit.

• 1.

### The free-body diagram below represents a 2000-kg elevator. What is the motion of the elevator if the tension in the cable is 1.96 ´ 104 N? (Assume 3 significant digits.)

• A.

The elevator might be accelerating upward.

• B.

The elevator might be accelerating downward.

• C.

The elevator must be at rest.

• D.

The elevator cannot be undergoing uniform motion.

• E.

The elevator must not be accelerating.

E. The elevator must not be accelerating.
Explanation
The free-body diagram shows that the only force acting on the elevator is the tension in the cable. Since the tension is equal to the weight of the elevator (mg), where m is the mass of the elevator and g is the acceleration due to gravity, the elevator is in equilibrium and not accelerating. Therefore, the correct answer is that the elevator must not be accelerating.

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• 2.

### Your "weight" is properly defined as

• A.

The amount of material of which you are composed

• B.

The gravitational force which Earth exerts on you

• C.

The gravitational force you exert on Earth

• D.

The force you exert on a set of bathroom scales

• E.

None of the above

B. The gravitational force which Earth exerts on you
Explanation
The correct answer is "the gravitational force which Earth exerts on you." This is because weight is a measure of the force of gravity acting on an object due to its mass. When we say someone's weight is 60 kg, it means that the Earth is exerting a force of 60 kg on that person.

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• 3.

### The value of "g" at the surface of Mars is 3.7 N/kg. How much would a 60.0-kg person weigh at an altitude above the Martian surface equivalent to the planet's radius?

• A.

2.2 x 102 N

• B.

1.6 x 102 N

• C.

1.1 x 102 N

• D.

56 N

• E.

28 N

D. 56 N
Explanation
At the surface of Mars, the value of "g" is given as 3.7 N/kg. The weight of a person is calculated by multiplying their mass by the value of "g". Since the person's mass is given as 60.0 kg, multiplying it by 3.7 N/kg gives a weight of 222 N. However, the question asks for the weight at an altitude above the Martian surface equivalent to the planet's radius. At this altitude, the person would be twice as far from the center of Mars, resulting in a decrease in the gravitational force. Therefore, the weight would be half of the original value, which is 56 N.

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• 4.

• A.

A

• B.

B

• C.

C

• D.

D

• E.

E

C. C
• 5.

### The gravitational field strength of Earth

• A.

Has a value of exactgly 9.8 N/kg [down] at all locations on its surface

• B.

Is greater at the equator than at the poles

• C.

Is smallest at the peak of Mount Everest, the highest elevation

• D.

Is largest at the deepest spot on the ocean floor

• E.

Is largest at the poles

E. Is largest at the poles
Explanation
The gravitational field strength of Earth is largest at the poles because the Earth is not a perfect sphere but rather an oblate spheroid, meaning it is slightly flattened at the poles and bulges at the equator. This bulge causes a greater concentration of mass at the equator, resulting in a slightly stronger gravitational field there. Conversely, the concentration of mass is slightly less at the poles, leading to a slightly weaker gravitational field compared to the equator. Therefore, the gravitational field strength is largest at the poles.

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• 6.

### If you weighed 112 N on the Moon where g = 1.6 N/kg, how much would you weigh on Earth?

• A.

1.1 x 102 N

• B.

1.7 x 104 N

• C.

6.9 x 102 N

• D.

1.1 x 104 N

• E.

6.9 x 103 N

C. 6.9 x 102 N
Explanation
The weight of an object is given by the formula W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. In this question, the weight on the Moon is given as 112 N and the acceleration due to gravity on the Moon is given as 1.6 N/kg. To find the weight on Earth, we can use the same formula and substitute the values. Therefore, the weight on Earth would be 6.9 x 102 N.

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• 7.

### According to Newton's law of universal gravitation, the gravitational force of attraction between two objects would be

• A.

Half as strong if they're moved twice as far apart

• B.

Twice as strong if they're moved half as far apart

• C.

Four times as strong if they're moved twice as far apart

• D.

Four times as strong if they're moved half as far apart

• E.

Twice as strong if they're moved twice as far apart

D. Four times as strong if they're moved half as far apart
Explanation
According to Newton's law of universal gravitation, the gravitational force of attraction between two objects is inversely proportional to the square of the distance between them. This means that if the distance between the objects is halved, the gravitational force will be four times stronger because (1/2)^2 = 1/4. Therefore, the answer "four times as strong if they're moved half as far apart" is correct.

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• 8.

### What would the gravitational field strength be on a planet with half Earth's mass and half its radius?

• A.

78.4 N/kg

• B.

39.2 N/kg

• C.

19.6 N/kg

• D.

9.8 N/kg

• E.

4.9 N/kg

C. 19.6 N/kg
Explanation
The gravitational field strength on a planet is determined by its mass and radius. If a planet has half the mass and half the radius of Earth, its gravitational field strength would be reduced by a factor of 4 (since the gravitational field strength is inversely proportional to the square of the radius). Therefore, the gravitational field strength on this planet would be 19.6 N/kg, which is half of Earth's gravitational field strength.

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• 9.

### Consider two planets, A and B. Planet A has half the mass and half the radius of planet B. The ratio of gA : gB would be

• A.

2 : 1

• B.

1 : 2

• C.

4 : 1

• D.

1 : 4

• E.

1 : 1

A. 2 : 1
Explanation
Planet A has half the mass and half the radius of planet B. The acceleration due to gravity (g) is directly proportional to the mass and inversely proportional to the square of the radius. Since planet A has half the mass and half the radius of planet B, the ratio of gA : gB would be (1/2) : (1/4), which simplifies to 2 : 1.

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• 10.

### Study the force system diagram pictured below and select the factor which would NOT influence the amount of kinetic friction.

• A.

Object's mass

• B.

Coefficient of kinetic friction

• C.

Normal force

• D.

Applied force

• E.

Gravitational field strength

D. Applied force
Explanation
The amount of kinetic friction is determined by factors such as the coefficient of kinetic friction, the normal force, and the gravitational field strength. The applied force, on the other hand, does not directly influence the amount of kinetic friction. Kinetic friction is the force that opposes the motion of an object, and it is dependent on the other factors listed in the question.

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• 11.

### The coefficient of friction stems from the

• A.

Nature of the two surfaces in contact

• B.

Mass of the object

• C.

Strength of the applied force

• D.

Strength of the normal force

• E.

Strength of the gravitational force

A. Nature of the two surfaces in contact
Explanation
The coefficient of friction is determined by the nature of the two surfaces in contact. Different materials have different levels of roughness and stickiness, which affect how easily they slide against each other. So, depending on the specific surfaces in contact, the coefficient of friction will vary. The mass of the object, strength of the applied force, strength of the normal force, and strength of the gravitational force do not directly determine the coefficient of friction.

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• 12.

### If the strength of the frictional force is equal to the applied force and oppositely directed, and assuming that all other forces may be ignored, the object

• A.

Must be at rest

• B.

Must be just about to move

• C.

May be at rest or moving at uniform velocity

• D.

Must be accelerating

• E.

Must be slowing down

C. May be at rest or moving at uniform velocity
Explanation
If the strength of the frictional force is equal to the applied force and oppositely directed, it means that the object experiences a balanced force. In this case, the object may be at rest if the applied force and frictional force cancel each other out, resulting in no net force acting on the object. Alternatively, the object may be moving at a constant velocity if the applied force and frictional force are equal in magnitude and opposite in direction, resulting in a net force of zero. Therefore, the object may be at rest or moving at uniform velocity in this scenario.

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• 13.

### A chalk brush sits on a metre stick as pictured in the diagram. As one end of the metre stick is elevated, the chalk brush eventually begins to slide. Why?

• A.

The coefficient of friction changes.

• B.

The gravitational force on the brush changes.

• C.

The normal force on the brush changes.

• D.

The gravitational force begins to act along the metre stick.

• E.

An applied force is created.

C. The normal force on the brush changes.
Explanation
When one end of the metre stick is elevated, the angle of the stick changes, causing the normal force on the brush to change. The normal force is the force exerted by a surface to support the weight of an object resting on it, and it acts perpendicular to the surface. As the angle of the stick changes, the normal force decreases, which reduces the friction between the brush and the stick. This decrease in friction allows the brush to slide. Therefore, the correct answer is that the normal force on the brush changes.

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• 14.

### A 425-g model rocket is accelerated upward at 86 m/s2 by its engine. What is the value of the force exerted by the engine on the rocket?

• A.

36.55 N [up]

• B.

41 N [down]

• C.

37 N [up]

• D.

32.55 N [up]

• E.

32 N [down]

A. 36.55 N [up]
Explanation
The force exerted by the engine on the rocket can be calculated using Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. In this case, the mass of the rocket is given as 425 grams, which is equal to 0.425 kg. The acceleration is given as 86 m/s^2. Multiplying these values together, we get a force of 36.55 N. Since the rocket is being accelerated upward, the force is directed upward as well.

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• 15.

### A 4.0-kg object, A, and a 2.0-kg object, B, are connected with a rope. A force is applied to another rope attached to the 2.0-kg object that pulls both A and B along a horizontal surface. Which of the following statements is true?

• A.

The force that B exerts on A is greater than the force that A exerts on B.

• B.

The force that A exerts on B is greater than the force that B exerts on A.

• C.

The force that B exerts on A is equal to the force that A exerts on B provided that the system slides with uniform motion.

• D.

The force that B exerts on A is equal to the force that A exerts on B regardless of the motion of the system.

• E.

The sum of the applied force and the force that B exerts on A is equal to the force that A exerts on B.

D. The force that B exerts on A is equal to the force that A exerts on B regardless of the motion of the system.
Explanation
In a system where two objects are connected with a rope, the force that one object exerts on the other is equal in magnitude but opposite in direction to the force that the other object exerts on the first. This is known as Newton's third law of motion. Therefore, the force that B exerts on A is equal to the force that A exerts on B regardless of the motion of the system.

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• 16.

### A 1.5-kg cart is pulled with a force of 7.3 N at an angle of 40° above the horizontal. If a kinetic friction force of 3.2 N acts against the motion, the cart’s acceleration along the horizontal surface will be

• A.

1.0 m/s2

• B.

1.6 m/s2

• C.

2.4 m/s2

• D.

2.7 m/s2

• E.

5.0 m/s2

B. 1.6 m/s2
Explanation
The cart is being pulled with a force of 7.3 N at an angle of 40° above the horizontal. To determine the acceleration, we need to resolve the force into its horizontal and vertical components. The horizontal component of the force is given by F*cos(40°) = 7.3 N * cos(40°) = 5.57 N.

The net force acting on the cart in the horizontal direction is the horizontal component of the applied force minus the kinetic friction force, so it is 5.57 N - 3.2 N = 2.37 N.

Using Newton's second law, F = ma, we can calculate the acceleration as a = F/m = 2.37 N / 1.5 kg = 1.58 m/s^2.

Rounding to the nearest tenth, the cart's acceleration along the horizontal surface will be 1.6 m/s^2.

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• 17.

### A 1.8-kg object is pulled along the floor with a force of 7.0 N acting horizontally. If the object accelerates at 2.4 m/s2, how much kinetic friction is acting?

• A.

30 N

• B.

11 N

• C.

8.3 N

• D.

7.8 N

• E.

2.7 N

E. 2.7 N
Explanation
The object is experiencing a horizontal force of 7.0 N and accelerating at 2.4 m/s^2. To determine the kinetic friction, we can use the equation Fnet = ma, where Fnet is the net force, m is the mass of the object, and a is the acceleration. Rearranging the equation, we have Fnet - ma = 0. The net force is the force applied (7.0 N) minus the force of kinetic friction (unknown). Therefore, 7.0 N - ma = 0. Plugging in the mass (1.8 kg) and acceleration (2.4 m/s^2), we can solve for the force of kinetic friction, which is 2.7 N.

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• 18.

### Which of the following statements concerning friction is true?

• A.

The frictional force always acts oppositely to the applied force.

• B.

For two given surfaces, the coefficient of static friction is generally greater than the coefficient of kinetic friction.

• C.

Friction is a force which is unavoidable and serves no practical purpose.

• D.

Two very highly-polished surfaces in contact with one another will have very little friction between them.

• E.

Friction always acts in the direction of motion.

B. For two given surfaces, the coefficient of static friction is generally greater than the coefficient of kinetic friction.
Explanation
The statement that the coefficient of static friction is generally greater than the coefficient of kinetic friction is true. The coefficient of static friction refers to the amount of force needed to overcome the static friction and set an object in motion. It is generally higher because it takes more force to overcome the initial resistance of an object at rest. Once the object is in motion, the coefficient of kinetic friction comes into play, which is generally lower because it requires less force to keep the object moving.

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• 19.

### If all other forces can be ignored and the strength of the frictional force is greater than the applied force and oppositely directed, the object

• A.

Could be speeding up or slowing down

• B.

Must be speeding up

• C.

Must be slowing down

• D.

Could be moving with uniform motion

• E.

Could be stopped

C. Must be slowing down
Explanation
If the strength of the frictional force is greater than the applied force and oppositely directed, it means that the frictional force is acting in the opposite direction to the motion of the object. This would cause the object to decelerate or slow down. Therefore, the object must be slowing down.

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• 20.

### The free-body diagram of a block being pushed up a rough ramp is best represented by

• A.

A

• B.

B

• C.

C

• D.

D

• E.

E

E. E
Explanation
The free-body diagram of a block being pushed up a rough ramp is best represented by option E. This is because option E shows the forces acting on the block, including the force of gravity pulling it down, the normal force perpendicular to the ramp, the frictional force opposing the motion, and the force applied to push the block up the ramp. The other options do not accurately represent all the forces acting on the block in this situation.

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Junho Song |Physics/Science Teacher |
Junho Song, a dedicated educator, specializes in teaching physics and science. Passionate about inspiring students in the fascinating world of scientific discovery and understanding.

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