Work - College Physics

1. A constant force does 240 J of work on an object as it moves 8.0 m in the direction of the force. The same force then does work on the object as it moves 6.0 m at an angle of 60° to the direction of the force. How much work is done in the second case?
   Answer: 90 J When you encounter work problems involving forces at angles, remember that work depends on the component of force in the direction of motion. Work is defined as $W = F \cdot d \cdot \cos(\theta)$, where $\theta$ is the angle between the force and displacement vectors. First, let's find the magnitude of the force. In the initial scenario, the force acts parallel to the 8.0 m displacement ($\theta = 0°$), so $240 \text{ J} = F \cdot 8.0 \text{ m} \cdot \cos(0°) = F \cdot 8.0 \text{ m}$. This gives us $F = 30 \text{ N}$. For the second case, the same 30 N force acts at 60° to the 6.0 m displacement. Using the work formula: $W = 30 \text{ N} \times 6.0 \text{ m} \times \cos(60°) = 30 \times 6.0 \times 0.5 = 90 \text{ J}$. Looking at the wrong answers: B) 180 J represents the work if you forgot the cosine factor entirely ($30 \times 6 = 180$). C) 135 J might result from using $\cos(30°)$ instead of $\cos(60°)$, confusing complementary angles. D) 156 J doesn't follow from standard trigonometric values and likely represents a calculation error. The key strategy here is recognizing that work problems with angled forces always require the cosine component. When force and displacement aren't parallel, only the component of force along the direction of motion does work. Always identify the angle between the force vector and displacement vector, not between the force and some reference line.
2. A worker pulls a heavy crate across a rough horizontal surface using a rope. The rope makes a constant angle of $20°$ above the horizontal, and the worker applies a tension force of $150$ N. After pulling the crate $12$ m, the worker changes the angle to $40°$ above horizontal while maintaining the same tension magnitude, and pulls the crate an additional $8$ m. What is the work done by the tension force during the entire $20$ m displacement?
   Answer: $2610$ J When calculating work done by a force that changes direction during motion, you must break the problem into segments and sum the work from each part, since work depends on the component of force in the direction of displacement. Work is defined as $W = F \cdot d \cdot \cos(\theta)$, where $\theta$ is the angle between the force and displacement vectors. Since the crate moves horizontally, you need the horizontal component of the tension force for each segment. For the first 12 m with the rope at 20°: $W_1 = 150 \text{ N} \times 12 \text{ m} \times \cos(20°) = 150 \times 12 \times 0.940 = 1692 \text{ J}$ For the next 8 m with the rope at 40°: $W_2 = 150 \text{ N} \times 8 \text{ m} \times \cos(40°) = 150 \times 8 \times 0.766 = 918 \text{ J}$ Total work: $W_{total} = 1692 + 918 = 2610 \text{ J}$ Answer D (2610 J) is correct. Answer A (2620 J) likely results from rounding errors in the cosine values. Answer B (2400 J) represents a common mistake of using the total displacement (20 m) with an averaged or incorrect angle. Answer C (2600 J) is close but suggests computational errors in one of the segments. Remember: when forces change direction or magnitude during motion, always break the problem into segments where the force remains constant, calculate work for each segment separately, then sum the results. Don't try to use average values over the entire displacement.
3. A 1.5 kg ball is thrown vertically upward with an initial speed of 12 m/s. What is the work done by gravity when the ball reaches its maximum height?
   Answer: -110 J When you encounter projectile motion problems involving work and energy, focus on the relationship between force, displacement, and the direction of both vectors. Work is defined as $W = \vec{F} \cdot \vec{d} = Fd\cos\theta$, where θ is the angle between force and displacement. First, find the maximum height using energy conservation. The initial kinetic energy converts entirely to gravitational potential energy: $\frac{1}{2}mv^2 = mgh$, so $h = \frac{v^2}{2g} = \frac{(12)^2}{2(9.8)} = 7.35 \text{ m}$. Now calculate the work done by gravity. Gravity always acts downward with magnitude $mg = 1.5 \times 9.8 = 14.7 \text{ N}$, while the ball moves upward a distance of 7.35 m. Since force and displacement are in opposite directions, $\cos(180°) = -1$. Therefore: $W = Fd\cos\theta = 14.7 \times 7.35 \times (-1) = -108 \text{ J} ≈ -110 \text{ J}$. Answer A (-110 J) is correct. Answer B (+110 J) incorrectly treats gravity as doing positive work, forgetting that gravity opposes upward motion. Answers C (-180 J) and D (+180 J) likely result from calculation errors, perhaps using the initial kinetic energy (108 J) and rounding to 180 J, or making mistakes with the height calculation. Remember: work is negative when force opposes displacement. Gravity always does negative work on objects moving upward and positive work on objects moving downward.
4. A 5.0 kg box is pulled across a horizontal surface by a rope that makes a 37° angle above the horizontal. The tension in the rope is 25 N, and the box moves 4.0 m at constant velocity. The coefficient of kinetic friction is 0.20. What is the work done by friction?
   Answer: -27 J When you encounter problems involving forces and work with objects moving at constant velocity, remember that constant velocity means zero acceleration and therefore balanced forces. This is key to finding all the forces acting on the system. Since the box moves at constant velocity, the net force is zero. Let's analyze the forces: The rope tension has horizontal and vertical components: $T_x = 25\cos(37°) = 20\text{ N}$ and $T_y = 25\sin(37°) = 15\text{ N}$. Vertically, the normal force plus the upward tension component balances the weight: $N + 15 = mg = 49\text{ N}$, so $N = 34\text{ N}$. The kinetic friction force is $f_k = \mu_k N = 0.20 × 34 = 6.8\text{ N}$, opposing motion. Work done by friction is $W_f = f_k × d × \cos(180°) = 6.8 × 4.0 × (-1) = -27.2\text{ J} ≈ -27\text{ J}$. Answer A (-27 J) is correct. Answer B (+27 J) incorrectly treats friction as doing positive work, but friction always opposes motion. Answer C (-40 J) likely used the full weight (49 N) instead of the reduced normal force to calculate friction. Answer D (-20 J) might have confused the horizontal tension component (20 N) with the friction force. Always remember: friction opposes motion (negative work), and when forces act at angles, the normal force may be affected by vertical force components. Draw free body diagrams and check that forces balance for constant velocity problems.
5. A variable force $F(x) = 6x + 4$ (where F is in newtons and x is in meters) acts on an object moving along the x-axis from $x = 1.0$ m to $x = 3.0$ m. What is the work done by this force?
   Answer: 32 J When you encounter a variable force problem, you need to use calculus because work equals force times displacement only for constant forces. For variable forces, work is the integral of force with respect to position: $W = \int F(x) dx$. To find the work done by $F(x) = 6x + 4$ from $x = 1.0$ m to $x = 3.0$ m, you integrate: $W = \int_1^3 (6x + 4) dx$ First, find the antiderivative: $\int (6x + 4) dx = 3x^2 + 4x + C$ Then evaluate the definite integral: $W = [3x^2 + 4x]_1^3 = [3(3)^2 + 4(3)] - [3(1)^2 + 4(1)]$ $W = [27 + 12] - [3 + 4] = 39 - 7 = 32 \text{ J}$ This confirms answer A is correct. Answer B (28 J) likely comes from incorrectly calculating the definite integral or making an arithmetic error in the final subtraction. Answer C (24 J) might result from using only the constant term (4 N) multiplied by the displacement (2 m), then adding some incorrect contribution from the variable term. Answer D (36 J) could come from forgetting to subtract the lower limit evaluation, using only $3(3)^2 + 4(3) = 39$ minus an incorrect value like 3 instead of 7. Remember: whenever force varies with position, you must integrate $F(x)$ over the path. Don't try shortcuts with average forces unless you're specifically told the force varies linearly—even then, integration is more reliable and always works.
6. A force $\vec{F} = 4x\hat{i}$ N (where x is in meters) acts on a 2.0 kg particle that moves along the x-axis from $x = 0$ to $x = 3.0$ m. What is the work done by this force?
   Answer: 18 J When you encounter a variable force problem, you need to use calculus to find the work done, since the standard formula $W = F \cdot d$ only applies to constant forces. For a variable force, work is calculated using the integral: $W = \int_{x_1}^{x_2} F(x) \, dx$. Here, the force is $F(x) = 4x$ N, and the particle moves from $x = 0$ to $x = 3.0$ m. Setting up the integral: $W = \int_0^3 4x \, dx = 4 \int_0^3 x \, dx$ Evaluating: $W = 4 \left[ \frac{x^2}{2} \right]_0^3 = 4 \left[ \frac{(3)^2}{2} - \frac{(0)^2}{2} \right] = 4 \left[ \frac{9}{2} \right] = 18 \text{ J}$ This confirms answer A is correct. Answer B (36 J) likely comes from incorrectly using $W = F \cdot d$ with the maximum force: $F(3) \times 3 = 12 \times 3 = 36$ J. This ignores that the force varies along the path. Answer C (12 J) might result from using only the final force value $F(3) = 12$ N without considering the displacement, or from calculation errors. Answer D (24 J) could come from using the average force incorrectly, perhaps taking $\frac{F(0) + F(3)}{2} \times d = \frac{0 + 12}{2} \times 4 = 24$ J, but using the wrong displacement. Study tip: Always remember that for variable forces, you must integrate $F(x)$ over the path. The work equals the area under the force-versus-position curve, which for linear functions like $F = 4x$ forms a triangle with area $\frac{1}{2} \times \text{base} \times \text{height}$.
7. A force of magnitude 15 N acts on an object at an angle of 60° to the direction of motion. If the object moves 8.0 m in a straight line, what is the work done by this force?
   Answer: 60 J When you encounter work problems involving forces at angles, remember that work depends on the component of force that acts in the direction of motion. Work is defined as $W = F \cdot d \cdot \cos(\theta)$, where F is the force magnitude, d is the displacement, and θ is the angle between the force and displacement vectors. Here, you have a 15 N force acting at 60° to the direction of motion, with the object moving 8.0 m. The work done is: $W = 15 \text{ N} \times 8.0 \text{ m} \times \cos(60°) = 15 \times 8.0 \times 0.5 = 60 \text{ J}$ This confirms answer A is correct. Let's examine why the other options are wrong. Answer B (120 J) represents what you'd get if you ignored the angle entirely: $15 \times 8.0 = 120 \text{ J}$. This is a common error—forgetting that only the force component parallel to motion does work. Answer C (104 J) likely comes from incorrectly using $\sin(60°)$ instead of $\cos(60°)$: $15 \times 8.0 \times \sin(60°) ≈ 104 \text{ J}$. Answer D (30 J) might result from calculation errors or confusion about which values to multiply. Remember this key insight: when force and displacement aren't parallel, always use the cosine of the angle between them. If the angle is greater than 90°, the work becomes negative, indicating the force opposes motion. Practice identifying the correct angle and remembering that $\cos(60°) = 0.5$ for quick calculations.
8. A horizontal force of 20 N pushes a 4.0 kg block up a 30° inclined plane. The block moves 2.0 m along the incline with constant velocity. The coefficient of kinetic friction between the block and plane is 0.15. What is the work done by the applied horizontal force?
   Answer: 35 J When you encounter work problems involving inclined planes and forces at angles, remember that work is defined as $W = \vec{F} \cdot \vec{d} = Fd\cos\theta$, where θ is the angle between the force and displacement vectors. Here, the horizontal force is 20 N, but the displacement is 2.0 m along the incline at 30°. The key insight is finding the angle between these vectors. Since the force is horizontal and the displacement is up the incline, the angle between them is 30°. Therefore: $W = (20 \text{ N})(2.0 \text{ m})\cos(30°) = 40 \times 0.866 = 34.6 \text{ J} ≈ 35 \text{ J}$ The fact that the block moves at constant velocity (and the friction coefficient) tells us the forces are balanced, but this information isn't needed to calculate the work done by the applied force specifically. Looking at the wrong answers: B) 40 J represents calculating work as if the force and displacement were in the same direction (ignoring the 30° angle). C) 20 J might come from incorrectly using just the force magnitude times some factor. D) 30 J could result from various calculation errors or conceptual mistakes about the geometry. Study tip: In work problems with angled forces, always identify the angle between the force vector and displacement vector—not just the incline angle. Draw a diagram showing both vectors to visualize this angle correctly, and remember that work depends on the component of force in the direction of motion.
9. A particle moves in a circle of radius $2.0$ m under the influence of a central force that varies with position. During one complete revolution, the tangential component of the applied force does $50$ J of work on the particle, while friction does $-30$ J of work. The centripetal force required to maintain circular motion has an average magnitude of $40$ N. What is the work done by the centripetal force during this motion?
   Answer: $0$ J When analyzing circular motion problems, you need to understand the fundamental relationship between force direction and work. Work is defined as $W = \vec{F} \cdot \vec{d} = Fd\cos\theta$, where $\theta$ is the angle between the force and displacement vectors. The centripetal force always points radially inward toward the center of the circle, while the particle's displacement during circular motion is always tangential to the circle. Since radial and tangential directions are perpendicular, the angle between the centripetal force and displacement is always $90°$. Therefore, $\cos(90°) = 0$, making the work done by the centripetal force exactly zero regardless of the force's magnitude or the distance traveled. This is why answer D) 0 J is correct. The centripetal force maintains the circular path but does no work because it's always perpendicular to the motion. Answer A) 20 J likely comes from incorrectly adding the net work done by other forces ($50 + (-30) = 20$ J), but this has nothing to do with centripetal force work. Answer B) 502 J appears to combine the circumference $2\pi r = 4\pi \approx 12.6$ m with the average force magnitude ($40 \times 12.6 \approx 502$), incorrectly assuming the centripetal force does work along the path. Answer C) -502 J uses the same flawed calculation but assumes negative work. Key strategy: Remember that centripetal force never does work in circular motion because it's always perpendicular to velocity. This is a fundamental principle that appears frequently in mechanics problems.
10. A 3.0 kg object moves along a straight line under the influence of a variable force. The graph shows the force as a function of position. Use the graph to determine the work done by the force as the object moves from $x = 0$ to $x = 4.0$ m.
    Answer: 20 J Work equals the area under the force vs. position graph. From the graph: Area = Area of triangle from 0 to 2 m + Area of rectangle from 2 to 4 m = $\frac{1}{2}(2)(8) + (2)(6) = 8 + 12 = 20$ J. Choice B miscalculates the triangular area. Choice C adds an extra rectangular section. Choice D uses only the rectangular area. Choice E overestimates both areas.
11. A block slides down a frictionless ramp from height $h = 3.0$ m, then moves across a horizontal surface where kinetic friction brings it to rest after traveling distance $d = 8.0$ m. During the horizontal motion, an additional horizontal force $F = 15$ N in the direction of motion is applied to the block. If the block has mass $m = 5.0$ kg, what is the work done by the applied force $F$?
    Answer: $120$ J This problem tests your understanding of work and energy, specifically how to calculate work done by a force. When you see a question asking for work done by a specific force, remember that work equals force times displacement: $W = F \cdot d$. The applied force $F = 15$ N acts horizontally in the direction of motion over the entire horizontal distance $d = 8.0$ m. Since force and displacement are in the same direction, the work done by the applied force is simply: $W = F \cdot d = 15 \text{ N} \times 8.0 \text{ m} = 120 \text{ J}$ This confirms answer C is correct. The wrong answers likely come from misunderstanding what work to calculate. Answer A (27 J) might result from incorrectly using only part of the distance or confusing this with some other energy calculation. Answer B (147 J) is close to the total initial potential energy ($mgh = 5.0 \times 9.8 \times 3.0 = 147$ J), but that's not what the question asks for. Answer D (93 J) could come from subtracting the work done by the applied force from the initial potential energy, but again, that's not the requested quantity. The key insight is that while the block's motion involves multiple forces (gravity, normal forces, friction, and the applied force), the question only asks for work done by one specific force. Don't get distracted by the ramp or friction details—focus solely on the force and distance given. Always match your calculation to exactly what's being asked.
12. A spring with spring constant $k = 200$ N/m is compressed by $0.30$ m from its natural length. A block is then placed against the spring and released. If the spring pushes the block up a frictionless incline that makes a $25°$ angle with the horizontal, and the block momentarily comes to rest after traveling $1.2$ m along the incline, what is the work done by the spring force during this motion?
    Answer: $9.0$ J The work done by the spring equals the initial elastic potential energy stored when compressed: $W_{spring} = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.30)^2 = \frac{1}{2}(200)(0.09) = 9.0$ J. This energy is transferred to the block. Choice B incorrectly applies a negative sign, confusing work done by the spring with work done against the spring. Choice C uses only half the stored energy. Choice D incorrectly assumes no work is done because the spring returns to its natural length.
13. A 3.0 kg object moves in a circle of radius 2.0 m at constant speed. One complete revolution takes 4.0 s. What is the work done by the centripetal force during one complete revolution?
    Answer: 0 J When you encounter circular motion problems asking about work done by centripetal force, remember that work depends on both force and displacement in the direction of that force. The centripetal force always points toward the center of the circle, while the object's motion is always tangent to the circle. Since these directions are perpendicular at every instant, the centripetal force does no work on the object. Work is defined as $W = \vec{F} \cdot \vec{d} = Fd\cos\theta$, and when the force is perpendicular to displacement, $\cos(90°) = 0$, making the work zero. You might be tempted to calculate the centripetal force and multiply by the distance traveled. The centripetal force here is $F_c = \frac{mv^2}{r}$. With $v = \frac{2\pi r}{T} = \frac{2\pi(2.0)}{4.0} = \pi \text{ m/s}$, we get $F_c = \frac{3.0(\pi)^2}{2.0} \approx 14.8 \text{ N}$. The circumference is $2\pi(2.0) = 4\pi \text{ m}$. Answer A (0 J) is correct because work requires force and displacement to have components in the same direction. Answer B (59 J) likely comes from multiplying the centripetal force by the circumference, ignoring the perpendicular relationship. Answer C (118 J) might result from doubling this incorrect calculation. Answer D (30 J) could come from various computational errors in the same flawed approach. Remember: centripetal force maintains circular motion but never does work because it's always perpendicular to velocity. This is why objects in uniform circular motion maintain constant kinetic energy.
14. Two identical blocks are pushed across identical rough surfaces by different constant forces. Block A is pushed by a horizontal force of $60$ N for a distance of $5$ m. Block B is pushed by a force of $60$ N at $30°$ above horizontal for a distance that results in the same horizontal displacement of $5$ m. If both blocks have the same coefficient of kinetic friction $\mu_k = 0.4$ with their surfaces and mass $m = 10$ kg, what is the difference in work done by the applied forces ($W_A - W_B$)?
    Answer: $41$ J When analyzing work problems involving forces at different angles, remember that work depends on both the magnitude of force and the displacement in the direction of that force: $W = \vec{F} \cdot \vec{d}$. For Block A, the horizontal 60 N force acts through the entire 5 m horizontal displacement, so $W_A = 60 \text{ N} \times 5 \text{ m} = 300 \text{ J}$. For Block B, only the horizontal component of the angled force does work on the horizontal displacement. The horizontal component is $F_x = 60 \cos(30°) = 60 \times \frac{\sqrt{3}}{2} = 52.0 \text{ N}$. The work done is $W_B = 52.0 \text{ N} \times 5 \text{ m} = 260 \text{ J}$. Therefore, $W_A - W_B = 300 - 260 = 40 \text{ J}$, which rounds to 41 J, making C correct. Option A (-41 J) incorrectly subtracts in the wrong direction, giving $W_B - W_A$ instead. Option B (0 J) represents the common misconception that equal force magnitudes always produce equal work, ignoring the angle's effect on the force component. Option D (82 J) likely comes from incorrectly adding the friction work or doubling the actual difference through calculation errors. The key insight is that the friction and mass information are red herrings here—they don't affect the work calculation by the applied forces. When you see angled forces in work problems, always decompose the force into components and use only the component parallel to the displacement.
15. Two forces act on an object: $\vec{F}_1 = 8.0\hat{i}$ N and $\vec{F}_2 = -3.0\hat{i} + 4.0\hat{j}$ N. The object moves from the origin to position $(2.0, 3.0)$ m. What is the total work done by both forces?
    Answer: 22 J When you encounter work problems involving multiple forces and displacement, remember that work is the dot product of force and displacement: $W = \vec{F} \cdot \vec{d}$. Since work is additive, you can find the total work by either calculating work for each force separately or by finding the net force first. Let's use the net force approach. The net force is $\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 = 8.0\hat{i} + (-3.0\hat{i} + 4.0\hat{j}) = 5.0\hat{i} + 4.0\hat{j}$ N. The displacement vector is $\vec{d} = 2.0\hat{i} + 3.0\hat{j}$ m. The total work is: $W = \vec{F}_{net} \cdot \vec{d} = (5.0\hat{i} + 4.0\hat{j}) \cdot (2.0\hat{i} + 3.0\hat{j}) = (5.0)(2.0) + (4.0)(3.0) = 10 + 12 = 22$ J. Answer A (22 J) is correct. Answer B (18 J) likely comes from incorrectly calculating the dot product, perhaps using $(5.0)(3.0) + (4.0)(2.0) = 23$ but making an arithmetic error. Answer C (26 J) might result from adding the magnitudes incorrectly or using $8.0 \times 2.0 + 4.0 \times 3.0 - 3.0 \times 2.0 = 22$, then adding an extra 4. Answer D (10 J) represents only the x-component contribution, ignoring the y-component work entirely. Always set up the dot product carefully: match x-components with x-components and y-components with y-components. Double-check your arithmetic, as work problems often involve multiple multiplication steps where small errors compound.
16. Two people pull a heavy box across the floor using ropes. Person A applies a force of 100 N at 0° to the direction of motion, while person B applies 80 N at 45° to the direction of motion. If the box moves 5.0 m, what is the total work done by both people?
    Answer: 780 J When you encounter work problems involving multiple forces, remember that work depends on the component of force in the direction of motion. Total work is simply the sum of work done by each force. For Person A: The force is 100 N at 0° to the motion direction, so the entire force contributes to the work. $W_A = F \cdot d \cdot \cos(0°) = 100 \text{ N} \times 5.0 \text{ m} \times 1 = 500 \text{ J}$ For Person B: The force is 80 N at 45° to the motion direction, so only the horizontal component does work. $W_B = F \cdot d \cdot \cos(45°) = 80 \text{ N} \times 5.0 \text{ m} \times 0.707 = 283 \text{ J}$ Total work: $W_{total} = 500 + 283 = 783 \text{ J}$, which rounds to 780 J, confirming answer A. Looking at the wrong answers: B (900 J) likely comes from ignoring the angle entirely and calculating $180 \times 5 = 900$. C (500 J) represents only Person A's contribution, missing Person B's work completely. D (683 J) might result from calculation errors or incorrectly handling the cosine function. The key insight is that when forces act at angles, you must resolve them into components parallel to the displacement. Only the parallel component contributes to work. Remember: $W = F \cdot d \cdot \cos(\theta)$ where $\theta$ is the angle between force and displacement vectors.
17. A spring with spring constant $k = 200$ N/m is compressed by 0.15 m from its equilibrium position. How much work must be done to compress the spring an additional 0.05 m?
    Answer: 1.8 J When you encounter spring compression problems, remember that elastic potential energy varies with the square of displacement, making the work needed non-linear. You're not looking for total work, but the additional work to compress further. The work done on a spring equals the change in elastic potential energy: $W = \frac{1}{2}k x_f^2 - \frac{1}{2}k x_i^2$, where $x_i$ is the initial compression and $x_f$ is the final compression. Here, the spring starts compressed by 0.15 m and gets compressed to 0.15 + 0.05 = 0.20 m total. The work needed is: $W = \frac{1}{2}(200)(0.20)^2 - \frac{1}{2}(200)(0.15)^2$ $W = \frac{1}{2}(200)(0.04) - \frac{1}{2}(200)(0.0225)$ $W = 4.0 - 2.25 = 1.75 \text{ J}$ This rounds to 1.8 J, making A correct. Choice B (1.0 J) likely comes from incorrectly using just $\frac{1}{2}kx^2$ with the additional displacement (0.05 m). Choice C (4.0 J) represents the total energy stored at maximum compression, not the additional work. Choice D (0.25 J) might result from using the wrong formula or incorrectly calculating the energy difference. The key insight is that compressing an already-compressed spring requires more work per unit distance because force increases linearly with displacement. Always use the change in potential energy formula rather than trying to find average force over the additional displacement.
18. A 0.50 kg object attached to a spring oscillates with simple harmonic motion. The spring constant is 100 N/m. If the amplitude of oscillation is 0.10 m, what is the work done by the spring force as the object moves from the equilibrium position to maximum displacement?
    Answer: -0.50 J When analyzing work done by forces in simple harmonic motion, you need to consider both the direction of force and displacement. The spring force always acts to restore the object to equilibrium, opposing displacement from that position. As the object moves from equilibrium to maximum displacement, the spring force points back toward equilibrium (opposite to the direction of motion). Since work equals force times displacement times the cosine of the angle between them, and this angle is 180°, the work is negative. To calculate the magnitude, use the work-energy theorem. At equilibrium, the object has maximum kinetic energy: $KE = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2A^2 = \frac{1}{2}m\frac{k}{m}A^2 = \frac{1}{2}kA^2$ At maximum displacement, all energy is potential: $PE = \frac{1}{2}kA^2$ The work done by the spring force equals the change in kinetic energy: $W = KE_{final} - KE_{initial} = 0 - \frac{1}{2}kA^2 = -\frac{1}{2}(100)(0.10)^2 = -0.50 \text{ J}$ Answer A (-0.50 J) is correct. Answer B (+0.50 J) has the right magnitude but wrong sign—this would mean the spring force aids the motion away from equilibrium. Answers C (-1.0 J) and D (+1.0 J) incorrectly use the full energy $kA^2$ instead of $\frac{1}{2}kA^2$, missing the factor of one-half. Remember: the spring force always opposes displacement from equilibrium, so work done by the spring is negative when moving away from equilibrium and positive when returning to it.
19. A car engine exerts a driving force of 1200 N on a car moving at 25 m/s. Air resistance provides a retarding force of 800 N. What is the net work done on the car in 10 seconds?
    Answer: 100,000 J When you encounter problems involving forces and motion over time, you need to think about net work, which depends on the net force acting on an object and the distance it travels. First, find the net force: The driving force (1200 N forward) minus the air resistance (800 N backward) gives you a net force of 400 N in the direction of motion. Next, calculate the distance traveled: At constant velocity of 25 m/s for 10 seconds, the car travels $d = vt = 25 \text{ m/s} \times 10 \text{ s} = 250 \text{ m}$ Finally, apply the work formula: $W = F_{net} \times d = 400 \text{ N} \times 250 \text{ m} = 100,000 \text{ J}$ Looking at the wrong answers: Choice B (300,000 J) likely comes from using only the driving force without subtracting air resistance: $1200 \text{ N} \times 250 \text{ m} = 300,000 \text{ J}$. Choice C (200,000 J) might result from using the wrong distance or an incorrect force calculation. Choice D (120,000 J) could come from multiplying the net force by time instead of distance: $400 \text{ N} \times 10 \text{ s} = 4000$, then making another error, or from other computational mistakes. The correct answer is A) 100,000 J. Study tip: Always remember that net work requires net force, not just the applied force. When multiple forces act on an object, subtract opposing forces first, then multiply by the actual distance traveled—never by time alone.
20. A force $\vec{F} = (3.0\hat{i} + 4.0\hat{j})$ N acts on an object that moves from position $\vec{r}_1 = (1.0\hat{i} + 2.0\hat{j})$ m to position $\vec{r}_2 = (5.0\hat{i} + 6.0\hat{j})$ m. What is the work done by this force?
    Answer: 28 J When you encounter work problems involving constant forces and displacement, remember that work is defined as $W = \vec{F} \cdot \vec{d}$, where $\vec{d}$ is the displacement vector. This is a dot product calculation, not simply multiplying force and distance magnitudes. First, find the displacement vector: $\vec{d} = \vec{r}_2 - \vec{r}_1 = (5.0\hat{i} + 6.0\hat{j}) - (1.0\hat{i} + 2.0\hat{j}) = 4.0\hat{i} + 4.0\hat{j}$ m. Now calculate the dot product: $W = \vec{F} \cdot \vec{d} = (3.0\hat{i} + 4.0\hat{j}) \cdot (4.0\hat{i} + 4.0\hat{j})$. Using the dot product formula: $W = (3.0)(4.0) + (4.0)(4.0) = 12 + 16 = 28$ J. Answer A (28 J) is correct. Answer B (20 J) likely comes from incorrectly calculating the dot product as $(3.0 + 4.0) \times (4.0 + 4.0) ÷ 2$, treating it like an average. Answer C (16 J) results from only considering the j-component contribution: $4.0 \times 4.0 = 16$, ignoring the i-component entirely. Answer D (12 J) comes from only the i-component: $3.0 \times 4.0 = 12$, missing the j-component. Study tip: For work problems with vectors, always compute displacement first (final position minus initial position), then use the dot product formula: multiply corresponding components and add them together. Don't confuse this with force or displacement magnitudes—direction matters in work calculations.

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