MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 4a Newtons Laws Free Body

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This deck focuses on 4a Newtons Laws Free Body, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is Newton’s third law stated for an interaction pair of forces?

Answer: $\vec{F}_{A\to B}=-\vec{F}_{B\to A}$ (equal magnitude, opposite direction). Newton's third law states that interaction forces between two objects are equal in magnitude but opposite in direction.

Flashcard 2: What is the SI unit of force, expressed in base SI units?

Answer: $1\ \text{N}=1\ \text{kg}\cdot\text{m}\cdot\text{s}^{-2}$. The newton derives from Newton's second law as the force accelerating 1 kg at 1 m/s².

Flashcard 3: What is the formula for the weight of an object of mass $m$ near Earth’s surface?

Answer: $\vec{W}=m\vec{g}$ (magnitude $W=mg$). Weight represents the gravitational force on an object, calculated as mass times gravitational acceleration.

Flashcard 4: What is the direction of the weight force vector $\vec{W}$ near Earth’s surface?

Answer: Downward, toward Earth’s center (along $-\hat{y}$ if up is $+\hat{y}$). Gravitational attraction directs weight toward Earth's center, opposite to the upward positive direction.

Flashcard 5: What is the definition of the normal force $\vec{N}$ in a free-body diagram?

Answer: Contact force exerted perpendicular to the surface on the object. Normal force prevents penetration by acting perpendicularly from the surface to the object in contact.

Flashcard 6: What is the maximum static friction magnitude in terms of $\mu_s$ and $N$?

Answer: $f_s\le \mu_s N$ (with $f_{s,\max}=\mu_s N$). Static friction can vary up to a maximum value given by the static friction coefficient times normal force.

Flashcard 7: What is the kinetic friction magnitude in terms of $\mu_k$ and $N$?

Answer: $f_k=\mu_k N$. Kinetic friction provides a constant force opposing sliding, proportional to normal force via the kinetic coefficient.

Flashcard 8: In a free-body diagram, which direction does friction act relative to the motion or impending motion?

Answer: Opposes relative motion (or impending relative motion) between surfaces. Friction acts to resist or prevent the tendency for relative movement between contacting surfaces.

Flashcard 9: What is the tension force $\vec{T}$ in an ideal (massless, inextensible) rope segment?

Answer: A pulling force along the rope; same magnitude throughout one ideal rope segment. Tension in an ideal rope is uniform, pulling equally at both ends along the rope's direction.

Flashcard 10: Which forces should be drawn on an object’s free-body diagram: internal forces or external forces?

Answer: External forces only (forces acting on the chosen object/system). Free-body diagrams depict only external forces on the system to analyze net force and motion.

Flashcard 11: Identify the correct condition for translational equilibrium using net force.

Answer: $\sum \vec{F}=\vec{0}$. Translational equilibrium occurs when net force is zero, resulting in constant velocity.

Flashcard 12: What does $\sum \vec{F}=\vec{0}$ imply about acceleration $\vec{a}$ for a constant-mass object?

Answer: $\vec{a}=\vec{0}$. From Newton's second law, zero net force implies zero acceleration for a constant-mass object.

Flashcard 13: What is the acceleration of a mass $m$ if the net force magnitude is $F_{\text{net}}$?

Answer: $a=\frac{F_{\text{net}}}{m}$. Newton's second law gives acceleration as net force magnitude divided by mass.

Flashcard 14: Find the net force in the $x$-direction if $F_1=10\ \text{N}$ right and $F_2=6\ \text{N}$ left.

Answer: $F_{\text{net},x}=4\ \text{N}$ right. Net force in a direction is the vector sum of component forces, here rightward minus leftward.

Flashcard 15: State Newton’s second law in vector form relating net force, mass, and acceleration.

Answer: $\sum \vec{F}=m\vec{a}$. Newton's second law quantifies that the net force on an object equals its mass times its acceleration vector.

Flashcard 16: What is Newton’s first law (law of inertia) stated in terms of net force and motion?

Answer: If $\sum \vec{F}=\vec{0}$, velocity is constant (rest or uniform straight-line motion). Newton's first law indicates that zero net force results in no acceleration, maintaining constant velocity including rest or uniform motion.

Flashcard 17: Find and correct the statement: “Action-reaction forces cancel because they act on the same object.”

Answer: Correct: action-reaction forces act on different objects, so they do not cancel on one FBD. Action-reaction pairs involve different objects, preventing cancellation in a single free-body diagram.

Flashcard 18: Identify the action-reaction partner to the normal force exerted by a table on a block.

Answer: Normal force exerted by the block on the table (equal magnitude, opposite direction). Newton's third law pairs the table's force on the block with the block's equal and opposite force on the table.

Flashcard 19: A hanging $3\ \text{kg}$ mass is at rest on a rope. What is the tension magnitude using $g=10\ \text{m}\cdot\text{s}^{-2}$?

Answer: $T=30\ \text{N}$. At rest, tension balances the weight to achieve vertical equilibrium.

Flashcard 20: A $10\ \text{kg}$ crate is pulled with $50\ \text{N}$ right; kinetic friction is $20\ \text{N}$ left. Find $a$.

Answer: $a=3\ \text{m}\cdot\text{s}^{-2}$ right. Net force is applied force minus opposing friction, then acceleration is net force over mass.

Flashcard 21: Find $f_k$ if $\mu_k=0.20$ and $N=60\ \text{N}$.

Answer: $f_k=12\ \text{N}$. Kinetic friction equals the kinetic coefficient multiplied by the normal force.

Flashcard 22: Find $f_{s,\max}$ if $\mu_s=0.40$ and $N=50\ \text{N}$.

Answer: $f_{s,\max}=20\ \text{N}$. Maximum static friction equals the static coefficient times the normal force magnitude.

Flashcard 23: A block rests on a horizontal table. If it is not accelerating vertically, what is $N$ in terms of $mg$?

Answer: $N=mg$. Vertical equilibrium requires normal force to counterbalance the downward weight.

Flashcard 24: What is the weight magnitude of a $5\ \text{kg}$ mass using $g=10\ \text{m}\cdot\text{s}^{-2}$?

Answer: $W=50\ \text{N}$. Weight magnitude is mass multiplied by gravitational acceleration near Earth's surface.

Flashcard 25: Find the acceleration if a $2\ \text{kg}$ object has a net force of $6\ \text{N}$ to the right.

Answer: $a=3\ \text{m}\cdot\text{s}^{-2}$ right. Acceleration equals net force divided by mass, in the direction of the net force.