Newton's Second Law - AP Physics 1

1. 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?
   Answer: $2.0\,\text{m/s}^2$ right 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.
2. 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?
   Answer: $1.25\,\text{m/s}^2$ right 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.
3. 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?
   Answer: $1.2\,\text{N}$ north 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.
4. 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?
   Answer: $4.0\,\text{kg}$ 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.
5. 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?
   Answer: $2.0\,\text{m/s}^2$ right 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.
6. 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?
   Answer: $1.4\,\text{m/s}^2$ 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.
7. A 2.5 kg cart experiences a horizontal net force of 5 N to the left due to a 12 N left pull and a 7 N right push. In the lab frame it accelerates left. What is its acceleration magnitude?
   Answer: $2\ \text{m/s}^2$ 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, with net force direction matching acceleration. The net force is 12 N left minus 7 N right, or 5 N left, so for 2.5 kg, acceleration is 2 m/s² left in the lab frame. This explains the leftward motion due to the unbalanced forces. A common distractor is choice C, 5 m/s², possibly from adding forces instead of subtracting opposites. Always determine net force direction by subtracting magnitudes of opposing forces before using a = F_net / m.
8. 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?
   Answer: $3\ \text{m/s}^2$ 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.
9. 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?
   Answer: $2\,\text{m/s}^2$ right 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.
10. 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?
    Answer: $4\,\text{m/s}^2$ right 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, $a_{\text{new}} = a_{\text{original}} \times \left( \frac{m_{\text{original}}}{m_{\text{new}}} \right)$.
11. 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?
    Answer: It would be cut in half. 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.
12. 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?
    Answer: $2.5\,\text{m/s}^2$ right 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.
13. A 1.5 kg cart on a track is acted on by two horizontal forces: 9 N right and 3 N left. In the lab frame, it accelerates right. What is the acceleration magnitude?
    Answer: $4\ \text{m/s}^2$ 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, where net force is the vector sum of all forces and determines acceleration direction. The net force here is 9 N right minus 3 N left, or 6 N right, so for 1.5 kg, acceleration is 4 m/s² right in the lab frame. This calculation connects the unbalanced forces to the rightward acceleration described. A common distractor is choice B, 6 m/s², possibly from dividing the larger force by mass without subtracting the opposing force. Always sum forces vectorially to find F_net before dividing by mass to find acceleration.
14. 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?
    Answer: $3\ \text{N}$ 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.
15. 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?
    Answer: $\tfrac{2}{3}\,\text{m/s}^2$ left 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.
16. 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?
    Answer: It is halved because the net force is unchanged but the mass doubles. 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.
17. 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?
    Answer: $12\,\text{N}$ right 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.
18. 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?
    Answer: $1.0\,\text{m/s}^2$ right 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.
19. 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?
    Answer: $12\ \text{N}$ to the right 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.
20. 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?
    Answer: $1\ \text{m/s}^2$ 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.

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