Newton's Second Law
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AP Physics 1 › Newton's Second Law
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?
$0.80,\text{m/s}^2$ right
$1.25,\text{m/s}^2$ right
$2.5,\text{m/s}^2$ right
$4.0,\text{m/s}^2$ 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.
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?
$2.0,\text{kg}$
$4.0,\text{kg}$
$10,\text{kg}$
$6.0,\text{kg}$
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.
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?
$2.0,\text{m/s}^2$ right
$3.0,\text{m/s}^2$ right
$1.0,\text{m/s}^2$ 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.
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?
$2.0,\text{m/s}^2$ right
$1.0,\text{m/s}^2$ right
$3.6,\text{m/s}^2$ right
$5.0,\text{m/s}^2$ 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.
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?
$2.4,\text{N}$ north
$0.30,\text{N}$ north
$0.60,\text{N}$ north
$1.2,\text{N}$ 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.
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?
$\tfrac{1}{3},\text{m/s}^2$ left
$6,\text{m/s}^2$ left
$\tfrac{2}{3},\text{m/s}^2$ left
$2,\text{m/s}^2$ 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.
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?
$0.58,\text{m/s}^2$
$2.4,\text{m/s}^2$
$3.4,\text{m/s}^2$
$1.4,\text{m/s}^2$
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.
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?
$2,\text{m/s}^2$ right
$3,\text{m/s}^2$ right
$6,\text{m/s}^2$ 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.
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
$2,\text{m/s}^2$ right
$4,\text{m/s}^2$ 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, $a_{\text{new}} = a_{\text{original}} \times \left( \frac{m_{\text{original}}}{m_{\text{new}}} \right)$.
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 is halved because the net force is unchanged but the mass doubles.
It becomes zero because friction opposes the motion.
It doubles because the applied force is unchanged.
It stays the same because the net force is unchanged.
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.