Physics Flashcards: Analyze Force Interactions Using Data

What this deck covers

This deck focuses on Analyze Force Interactions Using Data, giving you a quick way to review the definitions, rules, and examples that matter most for Physics.

How to use these flashcards

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 the formula for spring force magnitude using Hooke's law?

Answer: $F_s = kx$. Hooke's law: spring force is proportional to displacement.

Flashcard 2: What is the momentum formula for an object of mass $m$ moving at speed $v$?

Answer: $p=mv$. Momentum is the product of mass and velocity.

Flashcard 3: What is the formula for centripetal net force in uniform circular motion?

Answer: $F_c = \frac{mv^2}{r}$. Apply $F=ma$ with centripetal acceleration $a_c=\frac{v^2}{r}$.

Flashcard 4: What deck type best fits analyzing force interactions using data (recall, application, or both)?

Answer: both. This skill requires memorizing formulas and applying them to solve problems.

Flashcard 5: Identify the impulse if a constant net force of $6\,\text{N}$ acts for $0.50\,\text{s}$.

Answer: $3.0\,\text{N}\cdot\text{s}$. Using $J=F_{net}\Delta t=6\times 0.50=3.0\,\text{N}\cdot\text{s}$.

Flashcard 6: What is the formula relating net force, mass, and acceleration (Newton's 2nd law)?

Answer: $\sum F = ma$. Newton's second law states net force equals mass times acceleration.

Flashcard 7: Identify the acceleration if $F_{net}=18\,\text{N}$ and $m=6\,\text{kg}$ in the same direction.

Answer: $3\,\text{m/s}^2$. Using $a=\frac{F_{net}}{m}=\frac{18}{6}=3\,\text{m/s}^2$.

Flashcard 8: What is the maximum possible static friction force magnitude before slipping begins?

Answer: $f_s\le \mu_s N$. Static friction can vary up to coefficient times normal force.

Flashcard 9: What is the magnitude of the friction force for kinetic friction on a level surface?

Answer: $f_k=\mu_k N$. Kinetic friction equals coefficient times normal force.

Flashcard 10: Identify the normal force on a $10\,\text{kg}$ object resting on a level surface (no other vertical forces).

Answer: $N=mg=98\,\text{N}$. Normal force balances weight when at rest on level surface.

Flashcard 11: Identify the centripetal force if $m=0.50\ \text{kg}$, $v=6\ \text{m}\cdot\text{s}^{-1}$, and $r=3\ \text{m}$.

Answer: $6\ \text{N}$. Apply $F_c=\frac{mv^2}{r}$: $F_c=\frac{0.50\times6^2}{3}=\frac{18}{3}=6$.

Flashcard 12: Identify $f_k$ if $\mu_k=0.20$ and $N=50\,\text{N}$.

Answer: $10\,\text{N}$. Using $f_k=\mu_k N=0.20\times 50=10\,\text{N}$.

Flashcard 13: Identify $f_{s,max}$ if $\mu_s=0.40$ and $N=30\,\text{N}$.

Answer: $12\,\text{N}$. Using $f_{s,max}=\mu_s N=0.40\times 30=12\,\text{N}$.

Flashcard 14: Identify the average net force if momentum changes by $4.0\,\text{kg}\cdot\text{m/s}$ in $0.20\,\text{s}$.

Answer: $20\,\text{N}$. Using $F_{avg}=\frac{\Delta p}{\Delta t}=\frac{4.0}{0.20}=20\,\text{N}$.

Flashcard 15: Identify the weight of a $3.0\,\text{kg}$ object using $g=9.8\,\text{m/s}^2$.

Answer: $29.4\,\text{N}$. Using $W=mg=3.0\times 9.8=29.4\,\text{N}$.

Flashcard 16: What is the action-reaction rule for forces between two interacting objects?

Answer: Forces are equal in magnitude and opposite in direction. Newton's third law: every action has an equal and opposite reaction.

Flashcard 17: Identify $f_k$ if $\mu_k=0.20$ and $N=50\ \text{N}$.

Answer: $10\ \text{N}$. Apply $f_k=\mu_k N$: $f_k=0.20\times50=10\ \text{N}$.

Flashcard 18: Identify the net force if a $2.0\,\text{kg}$ cart accelerates at $4.0\,\text{m/s}^2$.

Answer: $8.0\,\text{N}$. Using $F_{net}=ma=2.0\times 4.0=8.0\,\text{N}$.

Flashcard 19: Identify the spring constant $k$ if $F=20\ \text{N}$ stretches a spring by $x=0.50\ \text{m}$.

Answer: $40\ \text{N}\cdot\text{m}^{-1}$. Rearrange Hooke's law: $k=\frac{F}{x}=\frac{20}{0.50}=40$.

Flashcard 20: Identify the spring force magnitude if $k=200\,\text{N/m}$ and $x=0.15\,\text{m}$.

Answer: $30\,\text{N}$. Using $F_s=kx=200\times 0.15=30\,\text{N}$.

Flashcard 21: Identify $f_s^{\max}$ if $\mu_s=0.40$ and $N=30\ \text{N}$.

Answer: $12\ \text{N}$. Apply $f_s^{\max}=\mu_s N$: $f_s^{\max}=0.40\times30=12\ \text{N}$.

Flashcard 22: What is the formula for centripetal acceleration for uniform circular motion?

Answer: $a_c = \frac{v^2}{r}$. Velocity squared divided by radius gives centripetal acceleration.

Flashcard 23: Identify the centripetal acceleration if $v=4\ \text{m}\cdot\text{s}^{-1}$ and $r=2\ \text{m}$.

Answer: $8\ \text{m}\cdot\text{s}^{-2}$. Apply $a_c=\frac{v^2}{r}$: $a_c=\frac{4^2}{2}=\frac{16}{2}=8$.

Flashcard 24: Identify the net force when forces of $12\,\text{N}$ right and $5\,\text{N}$ left act on an object.

Answer: $7\,\text{N}$ to the right. Net force is the vector sum: $12-5=7\,\text{N}$ rightward.

Flashcard 25: What is the formula for static friction magnitude when it is at its maximum value?

Answer: $f_s^{\max} = \mu_s N$. Maximum static friction equals coefficient times normal force.

Flashcard 26: What is the SI unit of force expressed in base units?

Answer: $1\ \text{N} = 1\ \text{kg}\cdot\text{m}\cdot\text{s}^{-2}$. Newton is derived from $F=ma$, giving kilogram-meter per second squared.

Flashcard 27: What is the spring force magnitude formula (Hooke's law) for displacement $x$?

Answer: $F_s=kx$. Spring force is proportional to displacement from equilibrium.

Flashcard 28: What is the impulse-momentum relation for constant net force over time interval $\Delta t$?

Answer: $J=F_{net}\Delta t=\Delta p$. Impulse equals force times time, which equals momentum change.

Flashcard 29: What is the formula for kinetic friction magnitude for a sliding object?

Answer: $f_k = \mu_k N$. Kinetic friction equals kinetic coefficient times normal force.

Flashcard 30: What is the weight (gravitational force) formula for an object of mass $m$ near Earth?

Answer: $W=mg$. Weight equals mass times gravitational acceleration ($g=9.8\,\text{m/s}^2$).

Flashcard 31: Identify the acceleration if $\sum F=10\ \text{N}$ and $m=2\ \text{kg}$.

Answer: $5\ \text{m}\cdot\text{s}^{-2}$. Rearrange $F=ma$ to get $a=\frac{F}{m}=\frac{10}{2}=5$.

Flashcard 32: What is the formula for weight near Earth's surface in terms of mass and $g$?

Answer: $W = mg$. Weight is gravitational force, calculated as mass times gravitational acceleration.

Flashcard 33: What is the formula for gravitational force between two masses separated by distance $r$?

Answer: $F_g = G\frac{m_1 m_2}{r^2}$. Newton's law of universal gravitation with inverse square dependence.

Flashcard 34: What is the formula relating net force, mass, and acceleration in one dimension?

Answer: $F_{net}=ma$. Newton's second law relates force to mass times acceleration.

Flashcard 35: Identify the mass if $\sum F=12\ \text{N}$ and $a=3\ \text{m}\cdot\text{s}^{-2}$.

Answer: $4\ \text{kg}$. Rearrange $F=ma$ to get $m=\frac{F}{a}=\frac{12}{3}=4$.

Flashcard 36: Identify the mass if $F_{net}=15\,\text{N}$ produces $a=3.0\,\text{m/s}^2$.

Answer: $5.0\,\text{kg}$. Using $m=\frac{F_{net}}{a}=\frac{15}{3.0}=5.0\,\text{kg}$.