Physics Flashcards: Apply Work Energy Theorem

What this deck covers

This deck focuses on Apply Work Energy Theorem, 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 work done by the normal force on an object moving along a surface it is perpendicular to?

Answer: $W_N=0$. Normal force is perpendicular to motion, so $\cos(90°) = 0$.

Flashcard 2: What is the spring potential energy for displacement $x$ from equilibrium?

Answer: $U_s=\frac{1}{2}kx^2$. Elastic potential energy is quadratic in displacement.

Flashcard 3: What is the energy form associated with gravity near Earth, using height $y$?

Answer: $U_g=mgy$. Gravitational potential energy increases with height.

Flashcard 4: Identify the formula for final speed when $W_{\text{net}}$ is known: start at $v_i$, mass $m$.

Answer: $v_f=\sqrt{v_i^2+\frac{2W_{\text{net}}}{m}}$. Derived from $W_{\text{net}} = \frac{1}{2}m(v_f^2 - v_i^2)$.

Flashcard 5: What is the formula for translational kinetic energy of a mass $m$ moving at speed $v$?

Answer: $K=\frac{1}{2}mv^2$. Kinetic energy depends on mass and the square of velocity.

Flashcard 6: What is the SI unit of work and energy?

Answer: $1\ \text{J}=1\ \text{N}\cdot\text{m}$. Joule equals newton-meter, the unit of work and energy.

Flashcard 7: Find $W_g$ for $m=5\ \text{kg}$ rising by $\Delta y=2\ \text{m}$ with $g=9.8\ \text{m/s}^2$.

Answer: $W_g=-98\ \text{J}$. $W_g = -mg\Delta y = -(5)(9.8)(2) = -98$ J for upward motion.

Flashcard 8: Find $W_{\text{net}}$ if $m=2\ \text{kg}$, $v_i=3\ \text{m/s}$, and $v_f=7\ \text{m/s}$.

Answer: $W_{\text{net}}=40\ \text{J}$. $\Delta K = \frac{1}{2}(2)(49-9) = 40$ J using work-energy theorem.

Flashcard 9: What is the work done by the gravitational force when an object changes height by $\Delta y$?

Answer: $W_g=-mg\Delta y$. Negative because gravity opposes upward displacement.

Flashcard 10: Find the work by gravity for $m=3\ \text{kg}$ rising by $\Delta y=2\ \text{m}$ with $g=9.8\ \text{m/s}^2$.

Answer: $-58.8\ \text{J}$. $W_g = -3 \times 9.8 \times 2 = -58.8$ J.

Flashcard 11: Find the work by kinetic friction if $f_k=6\ \text{N}$ acts over $d=4\ \text{m}$.

Answer: $W_f=-24\ \text{J}$. Friction opposes motion: $W_f = -(6)(4) = -24$ J.

Flashcard 12: What is the work done by a constant force $F$ over displacement $d$ at angle $\theta$?

Answer: $W=Fd\cos\theta$. $\theta$ is the angle between force and displacement vectors.

Flashcard 13: Find the work done by a $10\ \text{N}$ force over $3\ \text{m}$ when $\theta=60^\circ$.

Answer: $W=15\ \text{J}$. $W = 10 \times 3 \times \cos(60°) = 30 \times 0.5 = 15$ J.

Flashcard 14: What is the work done by a spring force when stretched from $x_1$ to $x_2$?

Answer: $W_s = \frac{1}{2}k(x_1^2-x_2^2)$. Spring work depends on initial and final position squares.

Flashcard 15: What is the Work-Energy Theorem written using net work and kinetic energy?

Answer: $W_{\text{net}}=\Delta K$. States that net work equals the change in kinetic energy.

Flashcard 16: What is the Work-Energy Theorem written as an equation for net work and kinetic energy?

Answer: $W_{net} = \Delta K$. Net work equals the change in kinetic energy.

Flashcard 17: What is the normal force work on an object moving along a level surface with no vertical displacement?

Answer: $W_N = 0$. Normal force perpendicular to motion does no work.

Flashcard 18: What is the SI unit of work and kinetic energy?

Answer: $1\ \text{J} = 1\ \text{N}\cdot\text{m}$. Joule equals newton-meter, the unit of energy.

Flashcard 19: Find $W$ when $F=8\ \text{N}$, $d=5\ \text{m}$, and $\theta=60^\circ$.

Answer: $20\ \text{J}$. $W = 8 \times 5 \times \cos(60°) = 40 \times 0.5 = 20$ J.

Flashcard 20: Find $v_f$ if $m=4\ \text{kg}$, $v_i=2\ \text{m/s}$, and $W_{net}=48\ \text{J}$.

Answer: $5\ \text{m/s}$. $48 = \frac{1}{2}(4)(v_f^2-4)$; solving gives $v_f = 5$ m/s.

Flashcard 21: What is the work done by kinetic friction of magnitude $f_k$ over distance $d$?

Answer: $W_f=-f_k d$. Friction always opposes motion, making work negative.

Flashcard 22: What is the work done by kinetic friction of magnitude $f_k$ over distance $d$?

Answer: $W_f = -f_k d$. Friction always opposes motion, doing negative work.

Flashcard 23: Which sign does work have when the force component is opposite the displacement direction?

Answer: $W<0$. Work is negative when force opposes motion ($\theta > 90°$).

Flashcard 24: Find $v_f$ if $m=4\ \text{kg}$, $v_i=0$, and $W_{\text{net}}=18\ \text{J}$.

Answer: $v_f=3\ \text{m/s}$. $18 = \frac{1}{2}(4)v_f^2$ gives $v_f^2 = 9$, so $v_f = 3$ m/s.

Flashcard 25: Find the work by friction if $f_k=6\ \text{N}$ acts over $d=4\ \text{m}$ in the direction opposite motion.

Answer: $-24\ \text{J}$. Friction opposes motion: $W_f = -6 \times 4 = -24$ J.

Flashcard 26: Find $W_{net}$ for $m=2\ \text{kg}$ when speed increases from $3\ \text{m/s}$ to $7\ \text{m/s}$.

Answer: $40\ \text{J}$. $W_{net} = \frac{1}{2}(2)(49-9) = 40$ J.

Flashcard 27: What is the gravitational work done on mass $m$ for a vertical change $\Delta y$ (upward positive)?

Answer: $W_g = -mg\Delta y$. Gravity does negative work when object rises.

Flashcard 28: Identify the work done by a force perpendicular to displacement (that is, $\theta = 90^\circ$).

Answer: $W = 0$. Perpendicular force does no work since $\cos(90°)=0$.

Flashcard 29: State the formula for translational kinetic energy of a mass $m$ moving at speed $v$.

Answer: $K = \frac{1}{2}mv^2$. Kinetic energy is half mass times velocity squared.

Flashcard 30: What is the net work if speed changes from $v_i$ to $v_f$ for mass $m$?

Answer: $W_{net} = \frac{1}{2}m(v_f^2-v_i^2)$. Apply work-energy theorem with initial and final speeds.

Flashcard 31: Find $W$ when $F=10\ \text{N}$, $d=3\ \text{m}$, and $\theta=0^\circ$.

Answer: $30\ \text{J}$. $W = 10 \times 3 \times \cos(0°) = 30$ J.

Flashcard 32: Which option gives the sign of work when $\theta = 180^\circ$ between $\vec{F}$ and $\vec{d}$?

Answer: $W<0$. Force opposite to displacement gives negative work.

Flashcard 33: Find $W_s$ for a spring with $k=200\ \text{N/m}$ moving from $x_i=0.30\ \text{m}$ to $x_f=0$.

Answer: $W_s=9\ \text{J}$. $W_s = \frac{1}{2}(200)(0.09-0) = 9$ J as spring relaxes.

Flashcard 34: What is the definition of work done by a constant force $\vec{F}$ over displacement $\vec{d}$?

Answer: $W = \vec{F}\cdot\vec{d} = Fd\cos\theta$. Work is the dot product of force and displacement vectors.

Flashcard 35: Identify the net work if an object moves in a circle at constant speed (uniform circular motion).

Answer: $W_{net}=0$. Constant speed means $\Delta K = 0$, so $W_{net} = 0$.