AP Physics 1 Quiz: Newtons Second Law

20 questions · exam conditions

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Newtons Second LawQuestion 1 of 20

In the lab frame, a $2.0\,\text{kg}$ cart on a level track is pulled right by a string with tension $8\,\text{N}$ . Kinetic friction on the cart is $3\,\text{N}$ left. The cart accelerates right. If the same forces act on a $4.0\,\text{kg}$ cart, what is the new acceleration?

2.5\,\text{m/s}^2

right

1.25\,\text{m/s}^2

right

4.0\,\text{m/s}^2

right

0.80\,\text{m/s}^2

right

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AP Physics 1 Quiz

AP Physics 1 Quiz: Newtons Second Law

Practice Newtons Second Law in AP Physics 1 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 AP Physics 1.

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

$2.5\,\text{m/s}^2$ right
$1.25\,\text{m/s}^2$ right (correct answer)
$4.0\,\text{m/s}^2$ right
$0.80\,\text{m/s}^2$ right

Explanation: This question assesses understanding of Newton's second law, which states that the net force on an object equals its mass times its acceleration (F_net = ma). The relationship F_net = ma implies that for a constant net force, acceleration is inversely proportional to mass, meaning doubling the mass halves the acceleration. In the original scenario, the net force is 8 N right minus 3 N left, resulting in 5 N right, so the 2.0 kg cart accelerates at 5/2 = 2.5 m/s² right. For the 4.0 kg cart with the same forces, the net force remains 5 N right, leading to an acceleration of 5/4 = 1.25 m/s² right as the increased mass resists change in motion more. A common distractor is 2.5 m/s² right, which might occur if one incorrectly assumes acceleration is independent of mass when forces are constant. A transferable strategy is to always calculate the net force as the vector sum of all individual forces before dividing by mass to find acceleration.

Question 2

In an inertial frame, a hockey puck of mass $0.20\,\text{kg}$ experiences a constant net force of $0.60\,\text{N}$ to the north and accelerates north. What net force would be required for the same puck to accelerate north at twice the rate?

$0.30\,\text{N}$ north
$0.60\,\text{N}$ north
$1.2\,\text{N}$ north (correct answer)
$2.4\,\text{N}$ north

Explanation: This question evaluates Newton's second law, F_net = ma, to find required force for desired acceleration. The law F_net = ma shows force proportional to acceleration for constant mass, so doubling acceleration requires doubling net force. Originally, a = 0.60 N / 0.20 kg = 3 m/s² north. For twice the rate (6 m/s²), needed F_net = 0.20 kg * 6 m/s² = 1.2 N north. The distractor 0.60 N north might come from thinking force unchanged for same direction. A transferable strategy is to rearrange F_net = m * desired a when scaling acceleration.

Question 3

In an inertial frame, a cart of mass $m$ experiences a constant net horizontal force $F_\text{net}$ to the left and accelerates left at $2\,\text{m/s}^2$ . If the cart's mass is tripled while $F_\text{net}$ remains the same, what is the new acceleration?

$6\,\text{m/s}^2$ left
$2\,\text{m/s}^2$ left
$\tfrac{2}{3}\,\text{m/s}^2$ left (correct answer)
$\tfrac{1}{3}\,\text{m/s}^2$ left

Explanation: This question examines Newton's second law, F_net = ma, focusing on how acceleration varies with mass under constant net force. The law F_net = ma shows acceleration inversely proportional to mass, so tripling mass reduces acceleration to one-third if F_net is unchanged. Originally, F_net = m * 2 m/s² left. With mass 3m, new acceleration is F_net / 3m = (m*2)/(3m) = 2/3 m/s² left, illustrating increased inertia. The distractor 2 m/s² left might come from assuming acceleration remains constant regardless of mass. A transferable strategy is to solve for unknown quantities by rearranging F_net = ma after identifying constants.

Question 4

A $1.5\,\text{kg}$ cart in the lab frame has a fan pushing it right with $6\,\text{N}$ while friction is $3\,\text{N}$ left; vertical forces cancel. The cart accelerates right. What is the cart's acceleration?

$6\,\text{m/s}^2$ right
$2\,\text{m/s}^2$ right (correct answer)
$3\,\text{m/s}^2$ right
$4\,\text{m/s}^2$ right

Explanation: This problem tests Newton's second law application with a fan-powered cart. The cart experiences 6 N right from the fan and 3 N left from friction, giving net force F_net = 6 N - 3 N = 3 N right. Applying F_net = ma: 3 N = (1.5 kg)(a), solving gives a = 2.0 m/s² right. Choice C (3 m/s²) incorrectly uses just the net force value without considering the mass. Remember that Newton's second law relates three quantities: net force, mass, and acceleration through F = ma.

Question 5

In the ground frame, a $3.0\,\text{kg}$ crate is pulled right by a rope with $18\,\text{N}$ . Kinetic friction is $6\,\text{N}$ left; vertical forces cancel. The crate accelerates right. What is the net force on the crate?

$24\,\text{N}$ right
$12\,\text{N}$ right (correct answer)
$6\,\text{N}$ left
$18\,\text{N}$ right

Explanation: This problem requires applying Newton's second law to find net force. The crate experiences an 18 N pull to the right and 6 N friction to the left. The net force is the vector sum: 18 N - 6 N = 12 N to the right. Since vertical forces cancel, this horizontal net force is the total net force on the crate. Choice A (24 N) incorrectly adds the forces instead of subtracting, while choice C gives the friction force with wrong direction. When finding net force, remember to subtract opposing forces and add forces in the same direction.

Question 6

A $8.0\,\text{kg}$ crate in the lab frame is pushed right with $28\,\text{N}$ . Kinetic friction is $12\,\text{N}$ left; weight and normal cancel. The crate accelerates right. What is the net force magnitude on the crate?

$40\,\text{N}$
$28\,\text{N}$
$16\,\text{N}$ (correct answer)
$12\,\text{N}$

Explanation: This problem tests finding net force magnitude using Newton's second law. The crate experiences 28 N right and 12 N friction left, giving net force F_net = 28 N - 12 N = 16 N right. The magnitude of this net force is 16 N. Choice A (40 N) incorrectly adds the forces instead of finding their vector sum, while choice B just gives the applied force. When finding net force, remember to subtract opposing forces, not add them.

Question 7

In an inertial frame, a $3.0\,\text{kg}$ cart is pulled right with $9\,\text{N}$ while a resistive force of $3\,\text{N}$ acts left, so it accelerates right. If the resistive force increases to $6\,\text{N}$ while the pull stays $9\,\text{N}$ , what is the new acceleration?

$1.0\,\text{m/s}^2$ right (correct answer)
$3.0\,\text{m/s}^2$ right
$2.0\,\text{m/s}^2$ right
$1.0\,\text{m/s}^2$ left

Explanation: This question assesses Newton's second law, F_net = ma, when one force changes while others remain constant. The relationship F_net = ma means acceleration changes proportionally with net force if mass is fixed. Originally, net force is 9 N right minus 3 N left = 6 N right, so a = 6/3 = 2 m/s² right for 3.0 kg. With resistive force now 6 N left, new net is 9-6 = 3 N right, giving a = 3/3 = 1.0 m/s² right. The distractor 3.0 m/s² right might occur if one subtracts incorrectly or ignores the change. A transferable strategy is to recalculate F_net whenever forces change, then find new a with a = F_net / m.

Question 8

In the ground frame, a $2.5\,\text{kg}$ block on a table is pulled right by $9\,\text{N}$ . Kinetic friction is $4\,\text{N}$ left; weight and normal cancel. The block accelerates right. What is the block's acceleration?

$5.0\,\text{m/s}^2$ right
$2.0\,\text{m/s}^2$ right (correct answer)
$1.0\,\text{m/s}^2$ right
$3.6\,\text{m/s}^2$ right

Explanation: This problem applies Newton's second law to find acceleration from forces. The block experiences 9 N right and 4 N friction left, so net force F_net = 9 N - 4 N = 5 N right. Using Newton's second law F_net = ma: 5 N = (2.5 kg)(a), which gives a = 2.0 m/s² right. Choice D (3.6 m/s²) might result from incorrectly using individual forces rather than net force. To solve these problems systematically, always find net force first by vector addition, then divide by mass.

Question 9

In the ground frame, a $6\,\text{kg}$ crate is pulled to the right by a horizontal force of $18\,\text{N}$ . Kinetic friction is $6\,\text{N}$ left, and the crate accelerates right. If the net force stays the same but the mass becomes $3\,\text{kg}$ , what is the acceleration?

$1\,\text{m/s}^2$ right
$4\,\text{m/s}^2$ right (correct answer)
$2\,\text{m/s}^2$ right
$12\,\text{m/s}^2$ right

Explanation: This question tests Newton's second law, $F_{\text{net}} = ma$ , exploring acceleration when mass changes but net force is constant. $F_{\text{net}} = ma$ indicates acceleration doubles if mass halves with fixed net force. Originally, net force is $18 \, \text{N}$ right minus $6 \, \text{N}$ left = $12 \, \text{N}$ right, $a = \frac{12}{6} = 2 \, \text{m/s}^2$ right. With mass $3 \, \text{kg}$ , new $a = \frac{12}{3} = 4 \, \text{m/s}^2$ right, showing reduced mass leads to greater acceleration. The distractor $2 \, \text{m/s}^2$ right might stem from using the original mass by mistake. A transferable strategy is to isolate variables: if $F_{\text{net}}$ constant, $a_{\text{new}} = a_{\text{original}} \times \left( \frac{m_{\text{original}}}{m_{\text{new}}} \right)$ .

Question 10

In an inertial frame, a box of mass $m$ is pushed right with $20\,\text{N}$ while friction is $5\,\text{N}$ left, so it accelerates right. If the box is replaced by one of mass $2m$ while the same forces act, how does the acceleration change?

It doubles because the applied force is unchanged.
It stays the same because the net force is unchanged.
It is halved because the net force is unchanged but the mass doubles. (correct answer)
It becomes zero because friction opposes the motion.

Explanation: This question tests Newton's second law, F_net = ma, highlighting how acceleration changes with mass when net force is constant. According to F_net = ma, acceleration is inversely proportional to mass, so doubling mass halves acceleration if net force stays the same. Originally, net force is 20 N right minus 5 N left = 15 N right, giving acceleration a = 15/m right. With mass 2m and same net force 15 N, new acceleration is 15/(2m) = (1/2)(15/m), halving the original value. The distractor 'It stays the same because the net force is unchanged' ignores the inverse relationship with mass. A transferable strategy is to identify if net force or mass changes, then apply a = F_net / m to compare accelerations.

Question 11

In the ground frame, a $10\,\text{kg}$ cart is pulled right by $40\,\text{N}$ while friction is $15\,\text{N}$ left; vertical forces cancel. The cart accelerates right. What is the cart's acceleration?

$2.5\,\text{m/s}^2$ right (correct answer)
$5.5\,\text{m/s}^2$ right
$4.0\,\text{m/s}^2$ right
$1.0\,\text{m/s}^2$ right

Explanation: This problem applies Newton's second law to a heavy cart. The cart experiences 40 N right and 15 N friction left, so net force F_net = 40 N - 15 N = 25 N right. Using F_net = ma: 25 N = (10 kg)(a), which gives a = 2.5 m/s² right. Choice B (5.5 m/s²) might come from using only partial forces or calculation errors. To avoid mistakes, write out the net force calculation explicitly before applying Newton's second law.

Question 12

A 1.0 kg puck on nearly frictionless ice is pushed right with 4 N while a fan exerts 1 N left. In the ice frame, the puck accelerates right. What is the net force magnitude on the puck?

$5\ \text{N}$
$4\ \text{N}$
$3\ \text{N}$ (correct answer)
$1\ \text{N}$

Explanation: This question assesses Newton's Second Law, which relates the net force on an object to its mass and acceleration. Newton's Second Law states that F_net = ma, but here we find net force directly from given forces. The net force is 4 N right minus 1 N left, or 3 N right, causing rightward acceleration in the ice frame. This magnitude is independent of mass since the question asks for net force, not acceleration. A common distractor is choice A, 5 N, possibly from adding the forces instead of subtracting. Always compute net force as the vector sum, considering directions, to apply or verify F_net = ma.

Question 13

In the ground frame, a cart is pushed right by $15\,\text{N}$ while friction is $9\,\text{N}$ left, producing a rightward acceleration of $1.5\,\text{m/s}^2$ . What is the cart's mass?

$2.0\,\text{kg}$
$4.0\,\text{kg}$ (correct answer)
$6.0\,\text{kg}$
$10\,\text{kg}$

Explanation: This question examines Newton's second law, F_net = ma, to determine mass from forces and acceleration. F_net = ma allows solving for m = F_net / a when net force and acceleration are known. Net force is 15 N right minus 9 N left = 6 N right. Given a = 1.5 m/s² right, mass m = 6 / 1.5 = 4.0 kg. The distractor 6.0 kg might arise from using gross force (15/1.5=10) minus something incorrectly. A transferable strategy is to compute F_net first, then use m = F_net / a for unknown mass.

Question 14

A 2.0 kg cart on a level, low-friction track is pulled right by a 10 N rope while friction pulls left with 4 N. In the ground frame, the cart accelerates right. What is the cart's acceleration magnitude?

$1\ \text{m/s}^2$
$2\ \text{m/s}^2$
$3\ \text{m/s}^2$ (correct answer)
$5\ \text{m/s}^2$

Explanation: This question assesses Newton's Second Law, which relates the net force on an object to its mass and acceleration. Newton's Second Law states that the net force F_net equals mass m times acceleration a, or F_net = ma, where the direction of acceleration matches the net force. In this scenario, the net force is the vector sum of the 10 N rightward pull minus the 4 N leftward friction, resulting in a 6 N net force to the right. This net force divided by the 2.0 kg mass gives an acceleration of 3 m/s² to the right, matching the described motion. A common distractor is choice B, 2 m/s², which might result from incorrectly using the mass as the net force or miscalculating the difference. Always identify all forces acting on the object and compute the net force vectorially before applying F_net = ma.

Question 15

A 4.0 kg cart on a track is acted on by 20 N right and 4 N left. In the ground frame, it accelerates right. If the net force stayed the same but the cart's mass doubled, what would happen to the acceleration?

It would double.
It would stay the same.
It would be cut in half. (correct answer)
It would become zero because forces balance.

Explanation: This question assesses Newton's Second Law, which relates the net force on an object to its mass and acceleration. Newton's Second Law states that a = F_net / m, showing inverse proportionality to mass for constant net force. Original net force is 20 N right minus 4 N left, or 16 N, giving 4 m/s² for 4.0 kg; doubling mass to 8.0 kg with same 16 N halves acceleration to 2 m/s². This connects the unchanged net force to reduced rightward acceleration in the ground frame. A common distractor is choice A, it would double, confusing the inverse relationship with direct proportionality. Always remember acceleration halves when mass doubles for constant net force, per a = F_net / m.

Question 16

In the ground frame, a $4.0\,\text{kg}$ cart is pulled right by $14\,\text{N}$ while friction is $6\,\text{N}$ left; vertical forces cancel. The cart accelerates right. What is the cart's acceleration?

$3.5\,\text{m/s}^2$ right
$2.0\,\text{m/s}^2$ right (correct answer)
$8.0\,\text{m/s}^2$ right
$0.50\,\text{m/s}^2$ right

Explanation: This problem applies Newton's second law to find acceleration. The cart experiences 14 N right and 6 N friction left, so net force F_net = 14 N - 6 N = 8 N right. Using Newton's second law F_net = ma: 8 N = (4.0 kg)(a), which gives a = 2.0 m/s² right. Choice C (8.0 m/s²) incorrectly uses the net force value as acceleration without dividing by mass. Always complete the calculation a = F_net/m to find the correct acceleration.

Question 17

In the ground frame, a $5.0\,\text{kg}$ object is acted on by $\vec F_1=12\,\text{N}$ east and $\vec F_2=5\,\text{N}$ west, and it accelerates east. What is the magnitude of its acceleration?

$3.4\,\text{m/s}^2$
$2.4\,\text{m/s}^2$
$1.4\,\text{m/s}^2$ (correct answer)
$0.58\,\text{m/s}^2$

Explanation: This question probes Newton's second law, F_net = ma, with opposing forces in one dimension. F_net = ma involves calculating the vector sum of forces to find net force, which divided by mass gives acceleration. Here, net force is 12 N east minus 5 N west = 7 N east for the 5.0 kg object. Acceleration magnitude is 7/5 = 1.4 m/s² east, linking the law to the object's eastward motion. The distractor 2.4 m/s² might result from adding forces without considering directions (12+5)/5. A transferable strategy is to assign positive/negative signs based on direction, sum for F_net, then compute a = F_net / m.

Question 18

In the ground frame, a 4.0 kg sled is pulled right by 14 N while friction is 6 N left, so it accelerates right. If the same net force acted on an 8.0 kg sled, what would the acceleration be?

$4\ \text{m/s}^2$
$2\ \text{m/s}^2$
$1\ \text{m/s}^2$ (correct answer)
$0.5\ \text{m/s}^2$

Explanation: This question assesses Newton's Second Law, which relates the net force on an object to its mass and acceleration. Newton's Second Law states that F_net = ma, meaning acceleration is directly proportional to net force and inversely proportional to mass. For the 4.0 kg sled, the net force is 14 N right minus 6 N left, or 8 N right, leading to 2 m/s² acceleration; applying the same 8 N net force to an 8.0 kg sled yields 1 m/s². This shows how doubling mass halves acceleration for constant net force, connecting to the rightward motion. A common distractor is choice B, 2 m/s², which might come from reusing the original acceleration without adjusting for mass change. Always solve for acceleration using a = F_net / m after determining if net force remains constant.

Question 19

In the lab frame, a 2.0 kg cart accelerates right at $3\ \text{m/s}^2$ while a 2 N friction force acts left. What is the magnitude of the applied rightward force on the cart?

$4\ \text{N}$
$6\ \text{N}$
$8\ \text{N}$ (correct answer)
$2\ \text{N}$

Explanation: This question assesses Newton's Second Law, which relates the net force on an object to its mass and acceleration. Newton's Second Law states that F_net = ma, so we can solve for unknown forces given acceleration. Here, F_net = 2.0 kg * 3 m/s² = 6 N right; with 2 N friction left, applied force = 6 N + 2 N = 8 N right. This balances the equation and explains the rightward acceleration in the lab frame. A common distractor is choice B, 6 N, which might ignore adding back the friction to find the applied force. Always use F_net = ma to find net force, then add opposing forces to solve for unknowns.

Question 20

A 3.0 kg box on a horizontal floor is pushed right with 18 N while kinetic friction is 6 N left. In the floor frame, the box accelerates right. What is the net force on the box?

$12\ \text{N}$ to the right (correct answer)
$18\ \text{N}$ to the right
$6\ \text{N}$ to the left
$24\ \text{N}$ to the right

Explanation: This question assesses Newton's Second Law, which relates the net force on an object to its mass and acceleration. Newton's Second Law states that the net force F_net equals mass m times acceleration a, or F_net = ma, where net force determines the direction and magnitude of acceleration. Here, the net force is the 18 N rightward push minus the 6 N leftward friction, yielding a 12 N net force to the right. This net force causes the 3.0 kg box to accelerate rightward in the floor frame, consistent with F_net = ma. A common distractor is choice B, 18 N to the right, which ignores the opposing friction force and uses only the applied force. Always calculate the net force by subtracting opposing forces to accurately apply F_net = ma.