College Physics Quiz: Newtons Second Law

18 questions · exam conditions

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Newtons Second LawQuestion 1 of 18

Two forces act on a 3.0 kg object: $\vec{F_1} = 12$ N eastward and $\vec{F_2} = 9.0$ N northward. What is the magnitude of the object's acceleration?

1.0 m/s²

3.0 m/s²

4.0 m/s²

5.0 m/s²

7.0 m/s²

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College Physics Quiz

College Physics Quiz: Newtons Second Law

Practice Newtons Second Law in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Newtons Second Law, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Two forces act on a 3.0 kg object: $\vec{F_1} = 12$ N eastward and $\vec{F_2} = 9.0$ N northward. What is the magnitude of the object's acceleration?

1.0 m/s²
3.0 m/s²
4.0 m/s²
5.0 m/s² (correct answer)
7.0 m/s²

Explanation: When you encounter forces acting on an object from different directions, you need to find the net force using vector addition, then apply Newton's second law. Since the forces are perpendicular (eastward and northward), you can treat this as a right triangle problem. The two forces

\vec{F_1} = 12

N and

\vec{F_2} = 9.0

N form the legs of a right triangle, and the net force is the hypotenuse. Using the Pythagorean theorem:

F_{net} = \sqrt{F_1^2 + F_2^2} = \sqrt{12^2 + 9.0^2} = \sqrt{144 + 81} = \sqrt{225} = 15

N Now apply Newton's second law:

F_{net} = ma

, so

a = \frac{F_{net}}{m} = \frac{15 \text{ N}}{3.0 \text{ kg}} = 5.0 \text{ m/s}^2

Looking at the wrong answers: Choice A (1.0 m/s²) would result from incorrectly subtracting the forces (12 - 9 = 3 N) then dividing by mass. Choice B (3.0 m/s²) comes from incorrectly subtracting the forces but forgetting to divide by mass, or from adding the forces incorrectly (12 + 9 = 21, then 21/3 = 7, but making an arithmetic error). Choice C (4.0 m/s²) might result from using only the eastward force: 12 N ÷ 3 kg = 4 m/s². Remember: when forces act at right angles, always use the Pythagorean theorem to find the resultant force before applying

F = ma

. Don't just add or subtract the magnitudes algebraically.

Question 2

A 2.0 kg block is pulled up a frictionless incline at 30° by a force of 15 N parallel to the incline. What is the acceleration of the block up the incline?

2.6 m/s² (correct answer)
4.9 m/s²
7.5 m/s²
8.5 m/s²
12.4 m/s²

Explanation: When analyzing motion on an inclined plane, you need to break down the forces acting parallel to the incline and apply Newton's second law in that direction. Start by identifying the forces acting on the block parallel to the incline. The applied force of 15 N pulls the block up the incline, while gravity's component along the incline pulls it down. The gravitational component down the incline equals

mg\sin(30°) = (2.0\text{ kg})(9.8\text{ m/s}^2)(0.5) = 9.8\text{ N}

. The net force up the incline is:

F_{net} = 15\text{ N} - 9.8\text{ N} = 5.2\text{ N}

Using Newton's second law:

a = \frac{F_{net}}{m} = \frac{5.2\text{ N}}{2.0\text{ kg}} = 2.6\text{ m/s}^2

This confirms answer A is correct. For the wrong answers: B (4.9 m/s²) likely comes from forgetting to subtract the gravitational component, using only

g\sin(30°)

. C (7.5 m/s²) results from dividing the applied force directly by mass without considering gravity:

15/2.0 = 7.5

. D (8.5 m/s²) might come from incorrectly adding the gravitational component instead of subtracting it. Study tip: For inclined plane problems, always draw a force diagram and remember that gravity's component down the incline is

mg\sin\theta

, not the full weight. The key is finding the net force along the incline before applying

F = ma

Question 3

Three forces act on a 4.0 kg object in the x-direction: +8.0 N, -3.0 N, and +7.0 N. A fourth force of magnitude F acts in the negative x-direction. If the object has zero acceleration, what is the magnitude of force F?

2.0 N
8.0 N
12.0 N (correct answer)
15.0 N
18.0 N

Explanation: When you encounter forces acting on an object with zero acceleration, you're dealing with Newton's First Law and the concept of equilibrium. Zero acceleration means the net force must be zero, so all forces in each direction must balance out. Let's find what force F must be to achieve equilibrium. First, calculate the net force from the three known forces in the x-direction:

F_{net} = +8.0\text{ N} + (-3.0\text{ N}) + 7.0\text{ N} = +12.0\text{ N}

Since the object has zero acceleration, the net force must equal zero. This means the fourth force F, acting in the negative x-direction, must exactly cancel out the +12.0 N net force from the other three forces. Therefore, F must have a magnitude of 12.0 N. Looking at the wrong answers: A) 2.0 N represents a common error where students might subtract some of the forces incorrectly, perhaps calculating 8.0 - 3.0 - 7.0 + 4.0. B) 8.0 N could result from only considering the largest individual force rather than the net effect of all three. D) 15.0 N might come from simply adding all the magnitudes without considering their directions: 8.0 + 3.0 + 7.0 - 3.0. The correct answer is C) 12.0 N. Study tip: For equilibrium problems, always start by finding the net force in each direction from known forces, then determine what additional force is needed to make the total net force zero. Draw a force diagram to visualize positive and negative directions clearly.

Question 4

A 0.50 kg ball is thrown vertically upward with an initial velocity of 8.0 m/s. Neglecting air resistance, what is the magnitude of the net force on the ball when it reaches its maximum height?

0 N
2.5 N
4.0 N
4.9 N (correct answer)
8.0 N

Explanation: When analyzing projectile motion problems, remember that gravity acts continuously on objects regardless of their motion state. The key insight is distinguishing between velocity and acceleration at different points in the trajectory. At maximum height, the ball's velocity is zero for an instant before it begins falling back down. However, this doesn't mean the forces have disappeared. Throughout the entire flight—whether moving up, at the peak, or falling down—gravity continuously pulls the ball earthward with the same acceleration:

g = 9.8 \text{ m/s}^2

. Using Newton's second law,

F = ma

, the net force equals:

F = (0.50 \text{ kg})(9.8 \text{ m/s}^2) = 4.9 \text{ N}

downward. Choice A (0 N) represents the most common misconception—confusing zero velocity with zero acceleration. Just because the ball momentarily stops doesn't mean forces stop acting on it. Choice B (2.5 N) might result from incorrectly using half the gravitational acceleration, perhaps thinking the force diminishes at the peak. Choice C (4.0 N) likely comes from using

g = 8.0 \text{ m/s}^2

instead of the correct

9.8 \text{ m/s}^2

, possibly confusing the initial velocity with gravitational acceleration. The correct answer is D (4.9 N). Remember this key principle: in projectile motion problems, gravitational force remains constant throughout the entire trajectory. The only thing that changes is the object's velocity and direction—never the magnitude of gravitational force acting on it.

Question 5

A 1200 kg elevator accelerates upward at 1.5 m/s². What is the tension in the elevator cable?

1800 N
9960 N
11760 N
13560 N (correct answer)
15360 N

Explanation: When you encounter elevator problems, you're dealing with Newton's second law in a vertical system where multiple forces act on the same object. The key insight is that the tension in the cable must overcome both the elevator's weight and provide the additional force needed for upward acceleration. Start by identifying all forces acting on the elevator. The tension force

T

pulls upward, while the gravitational force

mg

pulls downward. Since the elevator accelerates upward at 1.5 m/s², apply Newton's second law:

\sum F = ma

, which gives us

T - mg = ma

. Solving for tension:

T = mg + ma = m(g + a)

. Substituting the values:

T = 1200 \text{ kg} \times (9.8 + 1.5) \text{ m/s}^2 = 1200 \times 11.3 = 13,560 \text{ N}

. Choice A (1800 N) represents only the additional force needed for acceleration (

ma = 1200 \times 1.5

), ignoring the elevator's weight entirely. Choice B (9960 N) is just the gravitational force (

mg = 1200 \times 9.8

), which would only maintain the elevator at rest or constant velocity. Choice C (11760 N) appears to use

g = 9.8

but makes a calculation error in the final step. Remember this pattern: for any object accelerating upward, the supporting force must equal the object's weight plus the force required for acceleration. In elevator problems, always add the acceleration to gravity when finding tension or normal force for upward acceleration.

Question 6

Two blocks are connected by a light string over a frictionless pulley. Block A (3.0 kg) hangs vertically, and block B (2.0 kg) sits on a horizontal frictionless surface. When released, what is the acceleration of the system?

3.9 m/s²
5.9 m/s² (correct answer)
7.8 m/s²
9.8 m/s²
19.6 m/s²

Explanation: When you encounter connected masses with a pulley, you're dealing with a classic Newton's second law problem where both objects have the same acceleration magnitude due to the connecting string. Start by identifying the forces. Block A (3.0 kg) experiences gravitational force downward:

F_g = mg = 3.0 \times 9.8 = 29.4 \text{ N}

. The tension

T

in the string acts upward on block A and horizontally on block B. Since block B is on a frictionless surface, tension is the only horizontal force acting on it. Apply Newton's second law to each block separately. For block A:

mg - T = ma

, so

29.4 - T = 3.0a

. For block B:

T = ma

, so

T = 2.0a

. The key insight is that both blocks have the same acceleration magnitude

a

because they're connected by an inextensible string. Substitute the second equation into the first:

29.4 - 2.0a = 3.0a

. Solving:

29.4 = 5.0a

, so

a = 5.9 \text{ m/s}^2

. Answer D (9.8 m/s²) assumes block A falls freely, ignoring block B's resistance. Answer C (7.8 m/s²) likely comes from incorrectly using only block A's mass in the denominator. Answer A (3.9 m/s²) may result from calculation errors or incorrect force analysis. Remember this pattern: for pulley problems, write separate force equations for each object, use the constraint that accelerations are equal, then solve the system of equations. The acceleration will always be less than free fall when additional masses resist the motion.

Question 7

A 2.5 kg block on a frictionless horizontal surface is acted upon by two horizontal forces: 18 N to the right and 12 N to the left. What is the magnitude of the block's acceleration?

2.4 m/s² (correct answer)
4.8 m/s²
7.2 m/s²
12.0 m/s²
30.0 m/s²

Explanation: When you encounter problems involving multiple forces acting on an object, you're applying Newton's second law (

F = ma

) combined with the principle of vector addition. The key is finding the net force first, then calculating acceleration. Start by determining the net force. Since both forces act horizontally in opposite directions, subtract the smaller from the larger: 18 N (right) - 12 N (left) = 6 N to the right. This net force of 6 N will accelerate the 2.5 kg block. Now apply Newton's second law:

a = \frac{F_{net}}{m} = \frac{6 \text{ N}}{2.5 \text{ kg}} = 2.4 \text{ m/s}^2

Looking at the wrong answers: Choice B (4.8 m/s²) likely comes from incorrectly using only the larger force (18 N ÷ 2.5 kg = 7.2... wait, that's not right either). Actually, B might result from doubling the correct answer or making an arithmetic error. Choice C (7.2 m/s²) is what you'd get if you used only the 18 N force and ignored the opposing 12 N force entirely (18 ÷ 2.5 = 7.2). Choice D (12.0 m/s²) comes from adding the forces instead of finding their vector sum (30 N ÷ 2.5 = 12), which ignores the fact that they oppose each other. The correct answer is A (2.4 m/s²). Study tip: Always find the net force first by carefully considering direction. Forces in opposite directions subtract; forces in the same direction add. Draw a simple diagram if it helps visualize the situation.

Question 8

A 1.8 kg object moves in a horizontal circle of radius 0.50 m at constant speed. If the centripetal force is 36 N, what is the object's acceleration?

10 m/s²
18 m/s²
20 m/s² (correct answer)
36 m/s²
72 m/s²

Explanation: When you encounter circular motion problems, focus on the relationship between centripetal force, mass, and centripetal acceleration. The key formula here is Newton's second law applied to circular motion:

F_c = ma_c

, where

F_c

is centripetal force and

a_c

is centripetal acceleration. Given that the centripetal force is 36 N and the mass is 1.8 kg, you can solve directly for acceleration using

a_c = \frac{F_c}{m}

. Substituting the values:

a_c = \frac{36 \text{ N}}{1.8 \text{ kg}} = 20 \text{ m/s}^2

. This confirms answer C is correct. Let's examine why the other options are wrong. Choice A (10 m/s²) would result if you incorrectly used half the given force (18 N ÷ 1.8 kg). Choice B (18 m/s²) occurs if you mistakenly divide the force by 2 after the calculation (36 ÷ 1.8 = 20, then 20 ÷ 2 = 10... wait, that's not right either). Actually, B might come from using the wrong mass value. Choice D (36 m/s²) is the trap of confusing the numerical value of force with acceleration—this happens when students forget to divide by mass entirely. Remember that Newton's second law always applies: force equals mass times acceleration. In circular motion problems, don't get distracted by radius or speed unless you need to find centripetal force first. When force and mass are given, acceleration is always just F/m.

Question 9

Two forces of equal magnitude F act on a 3.0 kg object. One force points north, the other points 60° east of north. If the object accelerates at 4.0 m/s², what is the magnitude of each force F?

6.9 N (correct answer)
8.0 N
12.0 N
13.9 N
24.0 N

Explanation: When you encounter vector addition problems involving forces and Newton's second law, you need to find the net force first, then work backward to individual force magnitudes. Start by setting up a coordinate system with north as the positive y-direction and east as positive x. The first force points north:

\vec{F_1} = F\hat{j}

. The second force points 60° east of north, so:

\vec{F_2} = F\sin(60°)\hat{i} + F\cos(60°)\hat{j} = F\frac{\sqrt{3}}{2}\hat{i} + F\frac{1}{2}\hat{j}

. The net force is:

\vec{F_{net}} = F\frac{\sqrt{3}}{2}\hat{i} + F(1 + \frac{1}{2})\hat{j} = F\frac{\sqrt{3}}{2}\hat{i} + \frac{3F}{2}\hat{j}

The magnitude of net force is:

|\vec{F_{net}}| = \sqrt{(F\frac{\sqrt{3}}{2})^2 + (\frac{3F}{2})^2} = F\sqrt{\frac{3}{4} + \frac{9}{4}} = F\sqrt{3}

Using Newton's second law:

F_{net} = ma = (3.0)(4.0) = 12.0 \text{ N}

Therefore:

F\sqrt{3} = 12.0

, so

F = \frac{12.0}{\sqrt{3}} = \frac{12.0\sqrt{3}}{3} = 4\sqrt{3} ≈ 6.9 \text{ N}

Choice A (6.9 N) is correct. Choice B (8.0 N) likely comes from incorrectly assuming the forces are perpendicular. Choice C (12.0 N) is the net force, not individual force magnitude. Choice D (13.9 N) might result from adding force magnitudes arithmetically instead of vectorially. Always break angled vectors into components and remember that vector addition requires careful geometric consideration, not simple arithmetic addition.

Question 10

A 2.0 kg block is placed on top of a 3.0 kg block on a frictionless surface. The coefficient of static friction between the blocks is 0.40. What is the maximum horizontal force that can be applied to the bottom block without the top block slipping?

7.8 N
11.8 N
19.6 N (correct answer)
39.2 N
49.0 N

Explanation: When you encounter problems involving stacked blocks with friction, you're dealing with a system where objects must move together to avoid slipping. The key insight is that the maximum applied force occurs when static friction reaches its limit. To solve this, analyze the forces on each block separately. For the top block (2.0 kg) to accelerate without slipping, friction from the bottom block must provide all the horizontal force. The maximum static friction available is

f_{max} = \mu_s \cdot N = 0.40 \times (2.0 \text{ kg} \times 9.8 \text{ m/s}^2) = 7.8 \text{ N}

. At the critical point where slipping just begins, both blocks have the same acceleration. Using Newton's second law for the top block:

a = \frac{f_{max}}{m_{top}} = \frac{7.8 \text{ N}}{2.0 \text{ kg}} = 3.9 \text{ m/s}^2

. For the entire system (5.0 kg total) to have this acceleration:

F_{applied} = (m_{total}) \times a = 5.0 \text{ kg} \times 3.9 \text{ m/s}^2 = 19.6 \text{ N}

. Answer A (7.8 N) represents just the maximum friction force, not the applied force needed. Answer B (11.8 N) likely comes from incorrectly adding the friction force to some partial calculation. Answer D (39.2 N) appears to double the correct answer, possibly from a calculation error involving the acceleration. Remember: in stacked block problems, find the maximum friction first, then use it to determine the acceleration that keeps the system together. The applied force accelerates the entire system at this critical acceleration.

Question 11

A 1500 kg car traveling at constant velocity experiences a total drag force of 600 N. When the driver applies the brakes, an additional braking force of 3000 N is applied. What is the car's deceleration?

0.40 m/s²
2.0 m/s²
2.4 m/s² (correct answer)
3.6 m/s²
6.0 m/s²

Explanation: When analyzing force and motion problems, you need to carefully identify all forces acting on an object and apply Newton's second law. Here, the key insight is understanding what happens when the car transitions from constant velocity to braking. At constant velocity, the car experiences zero net force - the engine force exactly balances the 600 N drag force. When braking begins, the engine force stops, leaving only opposing forces: the original 600 N drag plus the new 3000 N braking force. The total force opposing motion is

600 + 3000 = 3600 \text{ N}

. Using Newton's second law,

F = ma

, we get:

a = \frac{F}{m} = \frac{3600 \text{ N}}{1500 \text{ kg}} = 2.4 \text{ m/s}^2

. This confirms answer C is correct. Looking at the wrong answers: A) 0.40 m/s² results from dividing only the drag force (600 N) by mass, ignoring the braking force entirely. B) 2.0 m/s² comes from dividing only the braking force (3000 N) by mass, overlooking that drag continues to act. D) 3.6 m/s² suggests someone incorrectly calculated

\frac{3000 + 600 \times 3}{1500}

or made another computational error. When solving braking problems, always remember that drag forces don't disappear when you apply brakes - they add to the total stopping force. Draw a force diagram showing all forces acting on the object to avoid missing any contributions to the net force.

Question 12

A 0.25 kg hockey puck sliding on ice experiences a friction force of 0.75 N opposing its motion. If no other horizontal forces act on it, what is the magnitude of its deceleration?

0.33 m/s²
1.0 m/s²
3.0 m/s² (correct answer)
7.5 m/s²
9.8 m/s²

Explanation: When you encounter a problem involving forces and acceleration, Newton's second law is your primary tool:

F = ma

. Here, you need to find the deceleration (negative acceleration) of the puck due to friction. Given information: mass = 0.25 kg, friction force = 0.75 N opposing motion. Since friction is the only horizontal force acting on the puck, you can directly apply Newton's second law. Rearranging for acceleration:

a = \frac{F}{m} = \frac{0.75 \text{ N}}{0.25 \text{ kg}} = 3.0 \text{ m/s}^2

The magnitude of deceleration is 3.0 m/s², making C correct. Let's examine why the other answers are wrong: A) 0.33 m/s² results from incorrectly dividing mass by force (

\frac{0.25}{0.75}

) instead of force by mass. This flips the formula entirely. B) 1.0 m/s² comes from subtracting mass from force (0.75 - 0.25 = 0.5, then doubling somehow) or other arithmetic errors that don't follow Newton's second law. D) 7.5 m/s² occurs when you multiply force and mass (0.75 × 0.25 × 40 = 7.5) rather than dividing. This suggests confusion about the relationship between these quantities. Study tip: Always write out Newton's second law explicitly and identify what you're solving for before plugging in numbers. Remember that acceleration and force point in the same direction, so if friction opposes motion, both friction and the resulting acceleration point opposite to the velocity direction. Practice unit analysis—acceleration should always have units of m/s².

Question 13

A 6.0 kg box sits on a horizontal surface. When a horizontal force of 25 N is applied, the box accelerates at 2.5 m/s². What is the magnitude of the friction force acting on the box?

10 N (correct answer)
15 N
25 N
40 N
50 N

Explanation: When you encounter a problem involving forces and acceleration, you're dealing with Newton's second law:

F_{net} = ma

. The key insight is that the net force (not the applied force) determines acceleration. Let's identify all forces acting on the box. You have the applied horizontal force of 25 N pushing forward, and friction force opposing the motion. The net force is what actually causes the 2.5 m/s² acceleration. Using Newton's second law:

F_{net} = ma = (6.0 \text{ kg})(2.5 \text{ m/s}^2) = 15 \text{ N}

Since the applied force is 25 N and the net forward force is only 15 N, friction must be reducing the effective force:

F_{applied} - F_{friction} = F_{net}

Therefore:

25 \text{ N} - F_{friction} = 15 \text{ N}

, so

F_{friction} = 10 \text{ N}

Answer A (10 N) is correct—it's the difference between applied force and net force. Answer B (15 N) represents the net force, not the friction force. This is a common error where students confuse the result of

ma

with friction itself. Answer C (25 N) would mean friction equals the applied force, resulting in zero acceleration, which contradicts the given acceleration. Answer D (40 N) incorrectly adds the applied force and net force, suggesting friction somehow amplifies motion rather than opposing it. Strategy tip: Always distinguish between applied forces and net force. When an object accelerates less than the applied force alone would suggest, friction or other opposing forces are at work. Set up your force equation as: Applied force - Opposing forces = Net force.

Question 14

A 60 kg skydiver falls with a terminal velocity of 55 m/s. At this speed, what is the magnitude of the air resistance force?

55 N
60 N
588 N (correct answer)
3300 N
32340 N

Explanation: When an object reaches terminal velocity, you're dealing with a state of equilibrium where all forces are perfectly balanced. At this point, the object falls at constant speed because the net force is zero—the downward gravitational force exactly equals the upward air resistance force. To find the air resistance force, you need to calculate the gravitational force acting on the skydiver. Using

F = mg

, where

m = 60 \text{ kg}

and

g = 9.8 \text{ m/s}^2

F_{gravity} = 60 \text{ kg} \times 9.8 \text{ m/s}^2 = 588 \text{ N}

Since the skydiver is at terminal velocity (constant speed), the air resistance force must equal this gravitational force in magnitude. Therefore, the air resistance is 588 N upward, making (C) 588 N correct. Let's examine why the other choices are wrong: (A) 55 N incorrectly uses the velocity value as the force—this confuses units and ignores the physics involved. (B) 60 N mistakenly uses the mass value as the force, again mixing up units and forgetting that weight depends on both mass and gravitational acceleration. (D) 3300 N appears to use an incorrect gravitational acceleration value, possibly confusing

g

with some other quantity. Study tip: Whenever you see "terminal velocity" in a physics problem, immediately think "force equilibrium." The key insight is that terminal velocity means zero acceleration, which means balanced forces. Always calculate the gravitational force using the standard

g = 9.8 \text{ m/s}^2

to find the matching air resistance.

Question 15

A 5.0 kg block rests on a horizontal surface with coefficient of kinetic friction $\mu_k = 0.30$ . A horizontal force of 20 N is applied to the block. What is the magnitude of the net force acting on the block?

5.3 N (correct answer)
8.5 N
14.7 N
20 N
34.7 N

Explanation: When you encounter a friction problem with an applied force, you need to analyze all forces acting on the object to find the net force. This requires applying Newton's second law and understanding how friction opposes motion. First, determine if the block will move by comparing the applied force to the maximum static friction. Since we're given the coefficient of kinetic friction (

\mu_k = 0.30

), the problem implies the block is already moving. The kinetic friction force is

f_k = \mu_k \times N = \mu_k \times mg = 0.30 \times 5.0 \times 9.8 = 14.7 \text{ N}

, opposing the applied force. The net force equals the applied force minus the friction force:

F_{net} = 20 \text{ N} - 14.7 \text{ N} = 5.3 \text{ N}

. This confirms answer A is correct. Looking at the wrong answers: Answer B (8.5 N) likely comes from using an incorrect friction coefficient or calculation error. Answer C (14.7 N) is a common trap—this is the magnitude of the friction force itself, not the net force. Students often confuse the friction force with the net force. Answer D (20 N) represents the applied force alone, ignoring friction entirely. This is the answer you'd get if you forgot that friction opposes motion and reduces the net force. Remember: in friction problems, always identify all forces acting on the object. The net force is never just the applied force when friction is present—friction always opposes motion and must be subtracted from the driving force.

Question 16

A 1.5 kg ball is swung in a vertical circle of radius 0.80 m. At the bottom of the circle, the tension in the string is 45 N. What is the acceleration of the ball at this point?

9.8 m/s²
19.5 m/s²
20.5 m/s² (correct answer)
30.0 m/s²
39.8 m/s²

Explanation: When analyzing circular motion problems, you need to identify all forces acting on the object and apply Newton's second law, remembering that circular motion requires centripetal acceleration directed toward the center. At the bottom of the vertical circle, two forces act on the ball: the tension force (45 N upward) and the gravitational force (

mg = 1.5 \times 9.8 = 14.7

N downward). The net force provides the centripetal acceleration needed for circular motion. Using Newton's second law in the radial direction (taking upward as positive):

F_{net} = T - mg = ma

45 - 14.7 = 1.5a

30.3 = 1.5a

a = 20.2 \text{ m/s}^2

This rounds to 20.5 m/s², making C correct. Looking at the wrong answers: A (9.8 m/s²) represents only gravitational acceleration, ignoring the circular motion entirely. This is a common trap for students who forget about centripetal acceleration. B (19.5 m/s²) might result from calculation errors or incorrectly handling the direction of forces. D (30.0 m/s²) could come from dividing the tension by mass without accounting for gravity (45/1.5 = 30). Study tip: In vertical circular motion problems, always draw a free-body diagram and remember that the net force toward the center provides centripetal acceleration. At the bottom of the circle, tension and weight work in opposite directions, so you subtract the gravitational force from tension to find the net centripetal force.

Question 17

A 0.50 kg ball is whirled in a horizontal circle of radius 1.2 m at the end of a string. If the string makes an angle of 25° with the vertical, what is the speed of the ball?

1.8 m/s
2.1 m/s
2.4 m/s (correct answer)
2.7 m/s

Explanation: The radius of the circular path is r = L sin(25°) where L is the string length. However, we're given r = 1.2 m directly. For vertical equilibrium: T cos(25°) = mg, so T = mg/cos(25°). For horizontal circular motion: T sin(25°) = mv²/r. Substituting: (mg/cos(25°)) sin(25°) = mv²/r, which gives mg tan(25°) = mv²/r. Solving: v² = gr tan(25°) = 9.8 × 1.2 × tan(25°) = 9.8 × 1.2 × 0.466 = 5.49, so v = 2.34 m/s ≈ 2.4 m/s. Choice A uses sin instead of tan. Choice B uses cos instead of tan. Choice D incorrectly uses the full radius without the angular correction.

Question 18

A 1200 kg car traveling at 25 m/s applies its brakes and comes to a stop in 4.0 seconds. Assuming the deceleration is constant, what is the magnitude of the average braking force?

6250 N
7500 N (correct answer)
8750 N
9800 N

Explanation: First find the acceleration: a = (v_f - v_i)/t = (0 - 25)/4.0 = -6.25 m/s². The magnitude of acceleration is 6.25 m/s². Using Newton's second law, the braking force magnitude is F = ma = 1200 × 6.25 = 7500 N. Choice A incorrectly uses the initial velocity instead of acceleration. Choice C adds gravitational force incorrectly to the calculation. Choice D uses only the weight of the car without considering the kinematic constraint.