Physics Midterm 1 Ultimate Study Guide

Dimensional Analysis and Unit Conversions

• Dimensional Consistency Analysis:

  • A physical equation is dimensionally correct if all terms separated by addition, subtraction, or equality signs possess identical fundamental dimensions.

  • Let $L$ represent length, $T$ represent time, and $M$ represent mass.

  • Equation i: $v_f = v_i + a t^2$

  • Left-hand side: $[v_f] = L T^{-1}$

  • Right-hand side terms: $[v_i] = L T^{-1}$, and $[a t^2] = (L T^{-2}) (T^2) = L$

  • Analysis: $L T^{-1} \neq L T^{-1} + L$. This equation is dimensionally incorrect.

  • Equation ii: $\Delta x = v_i t + \frac{1}{2} a t^2$

  • Left-hand side: $[\Delta x] = L$

  • Right-hand side terms: $[v_i t] = (L T^{-1}) T = L$, and $\left[\frac{1}{2} a t^2\right] = (L T^{-2}) T^2 = L$

  • Analysis: $L = L + L$. This equation is dimensionally correct.

  • Equation iii: $v_f^2 = v_i^2 + 2 a \Delta x$

  • Left-hand side: $[v_f^2] = (L T^{-1})^2 = L^2 T^{-2}$

  • Right-hand side terms: $[v_i^2] = L^2 T^{-2}$, and $[2 a \Delta x] = (L T^{-2}) L = L^2 T^{-2}$

  • Analysis: $L^2 T^{-2} = L^2 T^{-2} + L^2 T^{-2}$. This equation is dimensionally correct.

• Speed Limit Unit Conversion & Comparison:

  • Posted speed limit: $15\,\text{mph}$

  • Speedometer reading: $20\,\text{km/h}$

  • Conversion factor: $1\,\text{mph} \approx 1.60934\,\text{km/h}$ or $1\,\text{km/h} \approx 0.621371\,\text{mph}$

  • Converting $20\,\text{km/h}$ to miles per hour:

$20\,\text{km/h} \times \frac{1\,\text{mph}}{1.60934\,\text{km/h}} \approx 12.43\,\text{mph}$

• Converting $15\,\text{mph}$ to kilometers per hour:

$15\,\text{mph} \times 1.60934\,\text{km/h per mph} = 24.14\,\text{km/h}$

• Conclusion: Driving at $20\,\text{km/h}$ ($12.43\,\text{mph}$) is below the $15\,\text{mph}$ limit ($24.14\,\text{km/h}$). You are not speeding.

One-Dimensional Kinematics and Motion Analysis

• Kinematic Position Equation: $x(t) = 10 + 10t - 2t^2$

  • Comparing to standard kinematic form $x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$:

  • Initial position: $x_0 = 10\,\text{m}$

  • Initial velocity: $v_0 = 10\,\text{m/s}$

  • Acceleration parameter: $\frac{1}{2} a = -2\,\text{m/s}^2 \implies a = -4\,\text{m/s}^2$

  • Velocity function: $v(t) = \frac{dx}{dt} = 10 - 4t$

  • Acceleration function: $a(t) = \frac{dv}{dt} = -4\,\text{m/s}^2$

• Quantitative Tabulation of Motion ($t = 0\,\text{s}$ to $t = 4\,\text{s}$):

  • At $t = 0\,\text{s}$: $x(0) = 10\,\text{m}$, $v(0) = 10\,\text{m/s}$, $a(0) = -4\,\text{m/s}^2$

  • At $t = 1\,\text{s}$: $x(1) = 10 + 10(1) - 2(1)^2 = 18\,\text{m}$, $v(1) = 10 - 4(1) = 6\,\text{m/s}$, $a(1) = -4\,\text{m/s}^2$

  • At $t = 2\,\text{s}$: $x(2) = 10 + 10(2) - 2(2)^2 = 22\,\text{m}$, $v(2) = 10 - 4(2) = 2\,\text{m/s}$, $a(2) = -4\,\text{m/s}^2$

  • At $t = 2.5\,\text{s}$ (Turning point): $x(2.5) = 10 + 10(2.5) - 2(2.5)^2 = 22.5\,\text{m}$, $v(2.5) = 0\,\text{m/s}$, $a(2.5) = -4\,\text{m/s}^2$

  • At $t = 3\,\text{s}$: $x(3) = 10 + 10(3) - 2(3)^2 = 22\,\text{m}$, $v(3) = 10 - 4(3) = -2\,\text{m/s}$, $a(3) = -4\,\text{m/s}^2$

  • At $t = 4\,\text{s}$: $x(4) = 10 + 10(4) - 2(4)^2 = 18\,\text{m}$, $v(4) = 10 - 4(4) = -6\,\text{m/s}$, $a(4) = -4\,\text{m/s}^2$

• Verbal Description of Motion:

  • The object begins at position $x = 10\,\text{m}$ moving in the positive x-direction with an initial speed of $10\,\text{m/s}$.

  • It experiences a constant negative acceleration of $-4\,\text{m/s}^2$, causing it to slow down continuously as it moves rightward.

  • At $t = 2.5\,\text{s}$, the object comes to a instantaneous momentary rest at its maximum position of $x = 22.5\,\text{m}$.

  • After $t = 2.5\,\text{s}$, the object reverses direction, moving leftward (negative direction) with increasing speed.

Motion Diagrams

• Motion Diagram Principles:

  • Points represent positions at equal time intervals.

  • Velocity vectors (arrows) connect adjacent positions.

  • Acceleration vectors indicate rate and direction of velocity changes.

• Scenario Diagrams:

  • Object moving right at constant speed: Dots spaced equally from left to right; velocity vectors maintain equal length pointing right; acceleration vector is zero ($a = 0$).

  • Object moving right and speeding up at constant rate: Spacing between dots increases left to right; velocity vectors point right and grow progressively longer; acceleration vector points right ($a > 0$).

  • Object moving right and slowing down at constant rate: Spacing between dots decreases left to right; velocity vectors point right and grow progressively shorter; acceleration vector points left ($a < 0$).

  • Object moving left and speeding up at constant rate: Spacing between dots increases right to left; velocity vectors point left and grow progressively longer; acceleration vector points left ($a < 0$).

  • Object moving left and slowing down at constant rate: Spacing between dots decreases right to left; velocity vectors point left and grow progressively shorter; acceleration vector points right ($a > 0$).

Free-Fall Kinematics

• Hot-Air Balloon & Dropped Camera System:

  • Given Conditions:

  • Coordinate convention: Upward direction is positive ($+y$

  • Initial vertical position: $y_0 = 41\,\text{m}$

  • Final vertical position at ground level: $y = 0\,\text{m}$

  • Downward motion of balloon gives initial velocity of camera: $v_{0} = -2.3\,\text{m/s}$

  • Gravitational acceleration: $a = -g = -9.8\,\text{m/s}^2$

  • (a) Velocity Just Before Landing:

  • Kinematic equation:

$v_f^2 = v_0^2 + 2 a \Delta y$

- Substitution:

$v_f^2 = (-2.3\,\text{m/s})^2 + 2(-9.8\,\text{m/s}^2)(0\,\text{m} - 41\,\text{m})$

$v_f^2 = 5.29 + 803.6 = 808.89\,\text{m}^2/\text{s}^2$

$v_f = -\sqrt{808.89\,\text{m}^2/\text{s}^2} \approx -28.44\,\text{m/s} \approx -28\,\text{m/s}$

• (b) Time to Reach Ground:

  • Kinematic equation:

$v_f = v_0 + a t$

- Substitution:

$-28.44\,\text{m/s} = -2.3\,\text{m/s} + (-9.8\,\text{m/s}^2) t$

$-26.14\,\text{m/s} = -9.8\,\text{m/s}^2 t$

$t = \frac{-26.14}{-9.8} \approx 2.67\,\text{s} \approx 2.7\,\text{s}$

Multi-Phase Kinematics

• Taxi Motion Problem:

  • Phase 1: Constant Acceleration:

  • Initial velocity $v_0 = 0\,\text{m/s}$, final velocity $v_1 = 36.0\,\text{m/s}$, acceleration $a_1 = 2.00\,\text{m/s}^2$

  • Duration $t_1 = \frac{v_1 - v_0}{a_1} = \frac{36.0\,\text{m/s} - 0\,\text{m/s}}{2.00\,\text{m/s}^2} = 18.0\,\text{s}$

  • Displacement $\Delta x_1 = \frac{v_0 + v_1}{2} t_1 = \frac{0 + 36.0}{2} (18.0) = 324\,\text{m}$

  • Phase 2: Constant Velocity:

  • Velocity $v_1 = 36.0\,\text{m/s}$, duration $t_2 = 39.0\,\text{s}$

  • Displacement $\Delta x_2 = v_1 t_2 = (36.0\,\text{m/s}) (39.0\,\text{s}) = 1404\,\text{m}$

  • Phase 3: Uniform Deceleration:

  • Initial velocity $v_1 = 36.0\,\text{m/s}$, final velocity $v_2 = 0\,\text{m/s}$, duration $t_3 = 5.00\,\text{s}$

  • Displacement $\Delta x_3 = \frac{v_1 + v_2}{2} t_3 = \frac{36.0 + 0}{2} (5.00) = 90.0\,\text{m}$

  • i) Total Duration of Motion:

$t_{\text{total}} = t_1 + t_2 + t_3 = 18.0\,\text{s} + 39.0\,\text{s} + 5.00\,\text{s} = 62.0\,\text{s} \approx 62\,\text{s}$

• ii) Magnitude of Average Velocity:

  • Total displacement:

$\Delta x_{\text{total}} = \Delta x_1 + \Delta x_2 + \Delta x_3 = 324\,\text{m} + 1404\,\text{m} + 90.0\,\text{m} = 1818\,\text{m}$

- Average velocity magnitude:

$v_{\text{avg}} = \frac{\Delta x_{\text{total}}}{t_{\text{total}}} = \frac{1818\,\text{m}}{62.0\,\text{s}} \approx 29.32\,\text{m/s} \approx 29.3\,\text{m/s}$

Two-Dimensional Kinematics & Projectile Motion

• Rope Swing Jump into Swimming Hole:

  • Givens:

  • Launch speed: $v_0 = 2.25\,\text{m/s}$

  • Launch angle: $\theta = 35.00^\circ$ above horizontal

  • Time of flight: $t = 0.616\,\text{s}$

  • Gravitational acceleration: $g = 9.80\,\text{m/s}^2$

  • Vertical Velocity Component:

$v_{0y} = v_0 \sin(35.00^\circ) = 2.25 \times 0.573576 \approx 1.2905\,\text{m/s}$

• Height Determination:

  • Setting final height at water level $y = 0\,\text{m}$:

$y - y_0 = v_{0y} t - \frac{1}{2} g t^2$

$0 - y_0 = (1.2905\,\text{m/s})(0.616\,\text{s}) - \frac{1}{2}(9.80\,\text{m/s}^2)(0.616\,\text{s})^2$

$-y_0 = 0.7950\,\text{m} - 1.8593\,\text{m} = -1.0643\,\text{m}$

$y_0 = 1.06\,\text{m}$

• Football Kickoff Projectile Motion:

  • Givens:

  • Horizontal range: $R = 37\,\text{m}$

  • Launch angle: $\theta = 44^\circ$

  • Vertical displacement: $\Delta y = 0\,\text{m}$

  • Horizontal Range Formula:

$R = \frac{v_0^2 \sin(2\theta)}{g}$

• Solving for Initial Velocity $v_0$:

$37\,\text{m} = \frac{v_0^2 \sin(2 \times 44^\circ)}{9.80\,\text{m/s}^2} = \frac{v_0^2 \sin(88^\circ)}{9.80}$

$\sin(88^\circ) \approx 0.99939$

$v_0^2 = \frac{37 \times 9.80}{0.99939} \approx 362.82$

$v_0 = \sqrt{362.82} \approx 19.05\,\text{m/s} \approx 19.1\,\text{m/s}$

Vector Algebra and Dot Products

• Student Walk Vector Analysis:

  • Path segments: $A = 10\,\text{m}$, $B = 15\,\text{m}$, $C = 13\,\text{m}$, $D = 15\,\text{m}$

  • Unit vector notation: Cartesian displacement is expressed as $\mathbf{R} = R_x \mathbf{i} + R_y \mathbf{j}$

  • Components are summed separately along perpendicular axes:

$R_x = \sum V_x, \quad R_y = \sum V_y$

• Magnitude and angle of resultant displacement:

$R = \sqrt{R_x^2 + R_y^2}, \quad \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

• Vector Dot Product & Angle Between Vectors:

  • Givens:

  • $\mathbf{A} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}$

  • $\mathbf{B} = -\mathbf{i} + 2\mathbf{j} - \mathbf{k}$

  • (a) Scalar Product $\mathbf{A} \cdot \mathbf{B}$:

$\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z$

$\mathbf{A} \cdot \mathbf{B} = (2)(-1) + (3)(2) + (1)(-1) = -2 + 6 - 1 = 3$

• (b) Angle $\theta$ Between $\mathbf{A}$ and $\mathbf{B}$:

  • Magnitudes:

$|\mathbf{A}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \approx 3.7417$

$|\mathbf{B}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \approx 2.4495$

- Angle calculation:

$\cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{3}{\sqrt{14} \sqrt{6}} = \frac{3}{\sqrt{84}} \approx 0.327327$

$\theta = \cos^{-1}(0.327327) \approx 70.89^\circ \approx 71^\circ$

• Angle Between Two Vectors (Given Case):

  • Angle evaluated between specified vector directions: $102^\circ$

Newton's Laws and Statics

• Static Friction Ranking on Horizontal Surface:

  • Three identical boxes remain at rest under applied horizontal forces $F_1 = 30\,\text{N}$ and $F_2 = 20\,\text{N}$.

  • By Newton's First Law ($\sum F_x = 0$

$f_s = F_{\text{applied}}$

• Measured static friction magnitudes for the configuration: $1.2\,\text{N}$, $20\,\text{N}$, and $41\,\text{N}$.

• Ranking order of increasing force of static friction:

$\text{Box 3} < \text{Box 2} < \text{Box 1} \quad (1.2\,\text{N} < 20\,\text{N} < 41\,\text{N})$

• Suspended Flower Pot Statics:

  • Givens:

  • Pot mass: $m = 6.20\,\text{kg}$

  • Weight: $W = mg = (6.20\,\text{kg})(9.80\,\text{m/s}^2) = 60.76\,\text{N}$

  • Wire 1: Attached horizontally to wall (Tension $T_1$

  • Wire 2: Attached to ceiling at angle $\theta = 40.0^\circ$ above horizontal (Tension $T_2$

  • Equilibrium Equations:

  • Vertical direction:

$\sum F_y = T_2 \sin(40.0^\circ) - mg = 0$

$T_2 = \frac{60.76\,\text{N}}{\sin(40.0^\circ)} = \frac{60.76}{0.64279} \approx 94.525\,\text{N} \approx 94.6\,\text{N}$

- Horizontal direction:

$\sum F_x = T_2 \cos(40.0^\circ) - T_1 = 0$

$T_1 = T_2 \cos(40.0^\circ) = (94.525\,\text{N}) \cos(40.0^\circ) = 94.525 \times 0.76604 \approx 72.41\,\text{N} \approx 72.5\,\text{N}$

Dynamics on Inclines & Friction Systems

• Incline System Motion Data:

  • Displacement result up slope: $1.57\,\text{m}$

  • Incline acceleration result: $a = 6.21\,\text{m/s}^2$

  • Incline tension result: $T = 29.8\,\text{N}$

• Two-Body Incline and Hanging Mass System:

  • System Parameters:

  • Crate mass on incline: $m_1 = 300\,\text{kg}$

  • Suspended mass: $m_2 = 100\,\text{kg}$

  • Incline angle: $\theta = 30.00^\circ$

  • Kinetic friction coefficient: $\mu = 0.120$

  • **Forces on Incline Crate ($m_1$

  • Gravity parallel component: $m_1 g \sin(30.00^\circ) = (300)(9.80)(0.5) = 1470\,\text{N}$ (downward along slope)

  • Normal force: $F_N = m_1 g \cos(30.00^\circ) = (300)(9.80)(0.8660) = 2546.1\,\text{N}$

  • Friction force: $f_k = \mu F_N = 0.120 \times 2546.1 = 305.5\,\text{N}$

  • **Hanging Mass Force ($m_2$

  • Weight: $m_2 g = (100)(9.80) = 980\,\text{N}$

  • Net Dynamic Driving Force:

$F_{\text{net}} = m_1 g \sin(30.00^\circ) - m_2 g - f_k = 1470 - 980 - 305.5 = 184.5\,\text{N}$

• System Acceleration:

$a = \frac{F_{\text{net}}}{m_1 + m_2} = \frac{184.5\,\text{N}}{300\,\text{kg} + 100\,\text{kg}} = \frac{184.5}{400} \approx 0.46125\,\text{m/s}^2 \approx 0.462\,\text{m/s}^2$

• Cable Tension:

$T = m_2 (g + a) = (100\,\text{kg})(9.80\,\text{m/s}^2 + 0.462\,\text{m/s}^2) \approx 1027.2\,\text{N}$

Uniform Circular Motion

• Toy Airplane Flying in a Horizontal Circle:

  • System Parameters:

  • Mass of airplane: $m = 0.100\,\text{kg}$

  • Speed: $v = 3.50\,\text{m/s}$

  • Radius of horizontal circular path: $r = 1.50\,\text{m}$

  • d) Radial (Centripetal) Acceleration:

$a_r = \frac{v^2}{r} = \frac{(3.50\,\text{m/s})^2}{1.50\,\text{m}} = \frac{12.25}{1.50} \approx 8.1667\,\text{m/s}^2 \approx 8.17\,\text{m/s}^2$

• a) Angle of Inclination $\theta$ with Vertical:

  • Ratio of radial acceleration to gravitational acceleration:

$\tan(\theta) = \frac{a_r}{g} = \frac{8.1667\,\text{m/s}^2}{9.80\,\text{m/s}^2} \approx 0.8333$

$\theta = \tan^{-1}(0.8333) \approx 39.80^\circ$

• b) Tension in String:

$T \cos(\theta) = m g \implies T = \frac{m g}{\cos(39.80^\circ)} = \frac{(0.100\,\text{kg})(9.80\,\text{m/s}^2)}{\cos(39.80^\circ)} = \frac{0.980}{0.7683} \approx 1.28\,\text{N}$

• c) Components of Force Exerted by String:

  • Horizontal radial force component:

$T_x = T \sin(\theta) = m a_r = (0.100\,\text{kg})(8.1667\,\text{m/s}^2) \approx 0.819\,\text{N}$

- Vertical balancing force component:

$T_y = T \cos(\theta) = m g = (0.100\,\text{kg})(9.80\,\text{m/s}^2) = 0.983\,\text{N}$

- Vector formulation:

$\mathbf{F}_{string} = 0.819\mathbf{i} + 0.983\mathbf{j}$

Conceptual & Short-Answer Problems

• Question 1: Unaccelerated Object Forces

  • An object experiences zero acceleration ($a = 0$).

  • Analysis: By Newton's Second Law ($\sum \mathbf{F} = m \mathbf{a}$, zero acceleration requires the total net force to equal zero.

  • Evaluation of options:

  • (a) A single force acts on the object: Cannot be true (a single unbalanced force guarantees a non-zero acceleration).

  • (b) No forces act on the object: Can be true.

  • (c) Forces act, but cancel: Can be true.

• Question 2: Interplanetary Weight Comparison

  • Comparing 1 Newton of gold won on the Moon versus 1 Newton of gold won on Earth.

  • Analysis: Weight is given by $W = m g$, so mass is $m = \frac{W}{g}$. The Moon's surface gravity is approximately one-sixth of Earth's gravity ($g_{\text{Moon}} \approx \frac{1}{6} g_{\text{Earth}}$).

  • For equal weights ($W = 1\,\text{N}$), the mass of gold on the Moon is:

$m_{\text{Moon}} = \frac{1\,\text{N}}{g_{\text{Moon}}} \approx 6 \times \left(\frac{1\,\text{N}}{g_{\text{Earth}}}\right) = 6 m_{\text{Earth}}$

• Conclusion: Your friend is richer because 1 Newton of gold on the Moon represents about six times more mass (and intrinsic quantity) of gold than 1 Newton on Earth.

  • Question 3: Normal Force on an Incline

• A $100\,\text{g}$ ($0.100\,\text{kg}$) block rests on an incline of $\theta = 30^\circ$.

• Normal force formula:

$F_N = m g \cos(\theta) = (0.100\,\text{kg}) (9.80\,\text{m/s}^2) \cos(30^\circ) = (0.980\,\text{N})(0.8660) \approx 0.849\,\text{N}$

• Question 4: Work Done on a Pulled Incline Box

  • A box is pulled up a rough incline by a rope connected to a pulley.

  • Forces acting on the box:

  1. Tension force ($T$

  2. Kinetic friction force ($f_k$

  3. Gravitational force ($F_g$

  4. Normal force ($F_N$

  • Work evaluation: The Normal force acts perpendicular to the surface displacement ($\theta = 90^\circ$), so it does zero work ($W = F_N d \cos(90^\circ) = 0$). Tension, friction, and gravity all do non-zero work.

  • Answer: 3 forces are doing work on the box.

• Question 5: Speed Comparison with Equal Kinetic Energy

  • Car 1 has twice the mass of Car 2 ($m_1 = 2 m_2$), and both have equal kinetic energy ($K_1 = K_2$

  • Kinetic energy relation:

$\frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_2 v_2^2 \implies \frac{1}{2} (2 m_2) v_1^2 = \frac{1}{2} m_2 v_2^2$

$2 v_1^2 = v_2^2 \implies v_1 = \frac{v_2}{\sqrt{2}} \approx 0.707 v_2$

• Conclusion: Car 1 moves at $\frac{1}{\sqrt{2}}$ times the speed of Car 2 (or Car 2 moves $\sqrt{2}$ times faster than Car 1).

  • Question 6: Work Done on Skateboarder

• Speed increases from $2\,\text{m/s}$ to $3\,\text{m/s}$.

• Work-Energy Theorem: $W_{\text{ext}} = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 = \frac{1}{2} m (3^2 - 2^2) = \frac{5}{2} m > 0$

• Answer: a) Positive work was done.

  • Question 7: Work Comparison (Mimi vs. Joel)

• Mimi applied $10\,\text{N}$ over $3\,\text{m}$ in $10\,\text{s}$: $W_{\text{Mimi}} = F \times d = 10\,\text{N} \times 3\,\text{m} = 30\,\text{J}$

• Joel applied $10\,\text{N}$ over $3\,\text{m}$ in $1\,\text{min}$ ($60\,\text{s}$): $W_{\text{Joel}} = F \times d = 10\,\text{N} \times 3\,\text{m} = 30\,\text{J}$

• Conclusion: Both did equal work ($30\,\text{J}$). Mimi exerted greater power ($3\,\text{W}$ vs. $0.5\,\text{W}$), but work relies strictly on force and distance.

  • Question 8: Kilowatt-Hour Billing

• Kilowatt ($\text{kW}$) is a unit of power ($\text{J/s}$); hour ($\text{h}$) is a unit of time ($\text{s}$).

• Power multiplied by time yields energy ($\text{Power} \times \text{Time} = \text{Energy}$).

• Answer: a) Energy

  • Question 9: Negative Gravitational Potential Energy

• Answer: Yes. Gravitational potential energy ($U_g = m g y$) is defined relative to an arbitrarily chosen zero reference plane ($y = 0$). Any position below this reference level has a negative potential energy value.

  • Question 10: Path Independence of Potential Energy

• Two paths lead to the top of a hill: one steep and direct, the other twice as long but less steep.

• Analysis: Gravity is a conservative force. Potential energy change depends exclusively on initial and final vertical height ($\Delta U_g = m g \Delta h$).

• Conclusion: You gain the exact same potential energy regardless of path taken.

  • Question 11: Elastic Potential Energy in Springs

• Work required to stretch a spring is given by $W = \frac{1}{2} k x^2$

• Ratio comparing stretching $2\,\text{cm}$ versus $1\,\text{cm}$:

$\frac{W_{2\text{cm}}}{W_{1\text{cm}}} = \frac{\frac{1}{2} k (2\,\text{cm})^2}{\frac{1}{2} k (1\,\text{cm})^2} = \frac{4}{1} = 4$

• Conclusion: Stretching a spring $2\,\text{cm}$ requires 4 times as much work as stretching it $1\,\text{cm}$.