Work - AP Physics C Mechanics

1. A box of mass $m$ is pulled across a horizontal surface with coefficient of kinetic friction $\mu_k$ by a rope making angle $\alpha$ above the horizontal. The applied force has magnitude $F$ and the box moves at constant velocity for a distance $d$. What is the work done by the normal force from the surface on the box?
   Answer: $0$ The normal force is always perpendicular to the surface, pointing vertically upward with magnitude $N = mg - F\sin\alpha$. The displacement is horizontal. Since the normal force is perpendicular to the displacement, the work done is $W = \vec{N} \cdot \vec{d} = 0$. Choice B incorrectly multiplies the normal force magnitude by distance. Choice C gives the work done by friction. Choice D gives the work done by the applied force component.
2. A particle moves along the x-axis under the influence of a force given by $F(x) = kx^3$, where $k$ is a constant. What is the work done by this force as the particle moves from $x = 0$ to $x = L$?
   Answer: $\frac{1}{4}kL^4$ Work done by a variable force is the integral of the force with respect to displacement. $W = \int_{x_i}^{x_f} F(x) dx$. In this case, $W = \int_{0}^{L} kx^3 dx = k \left[ \frac{x^4}{4} \right]_{0}^{L} = k \left( \frac{L^4}{4} - 0 \right) = \frac{1}{4}kL^4$.
3. A $3.0\,\text{kg}$ block is pulled up a frictionless incline of angle $35^\circ$ by a constant force $F=25\,\text{N}$ applied at an angle of $15^\circ$ above the incline surface. The block moves $d=2.0\,\text{m}$ along the incline. The angle between the applied force and the displacement is $\theta=15^\circ$. Based on the scenario described above, calculate the work done in moving the object up the incline using $W=Fd\cos(\theta)$.
   Answer: $48\,\text{J}$ This question tests understanding of work when force is applied at an angle above an inclined surface. Work is calculated as W = Fd cos(θ), where θ is the angle between force and displacement. A 25 N force is applied 15° above the incline surface, moving the block 2.0 m along the incline, so the angle between force and displacement is 15°. Choice C is correct: W = (25 N)(2.0 m)cos(15°) = 50 × 0.966 = 48.3 J ≈ 48 J. Choice D incorrectly assigns a negative value, which would occur if force opposed motion. When teaching incline problems with angled forces, carefully distinguish between angles relative to horizontal, relative to incline, and between force-displacement vectors. Use clear diagrams showing all relevant angles to prevent confusion.
4. A force $\vec{F} = (2x\hat{i} + 3y^2\hat{j} - 4z^3\hat{k})$ N acts on a particle. What is the work done by this force as the particle moves from the origin (0, 0, 0) to the point (1, 1, 1) m? The force is conservative.
   Answer: 1 J Since the force is conservative, the work done is independent of the path and is equal to the negative change in potential energy, $W = -\Delta U$. The potential energy function is found by integrating the components of the force: $U(x,y,z) = -\int F_x dx - \int F_y dy - \int F_z dz = -x^2 - y^3 + z^4$. The work done is $W = -[U(1,1,1) - U(0,0,0)] = -[(-1^2 - 1^3 + 1^4) - (0)] = -[-1 - 1 + 1] = -[-1] = 1 \text{ J}$.
5. In a mechanics lab, a $5.0\,\text{kg}$ block is pulled across a frictionless horizontal track by a constant force of magnitude $F=18\,\text{N}$ applied at an angle $\theta=25^\circ$ above the horizontal. The block moves a horizontal displacement of $d=4.0\,\text{m}$. Based on the scenario described above, how much work is done by the force in pulling the object? Use $W=Fd\cos(\theta)$.
   Answer: $65\,\text{J}$ This question tests understanding of work in physics, specifically the calculation of work done by a force applied at an angle. Work is defined as the product of force and displacement in the direction of the force, calculated using W = Fd cos(θ). In this scenario, you are given a force of 18 N applied at an angle of 25° above the horizontal, and a horizontal displacement of 4.0 m. Choice C is correct because it accurately applies the work formula: W = (18 N)(4.0 m)cos(25°) = 72 × 0.906 = 65.2 J ≈ 65 J. Choice A incorrectly calculates 18 × 4 = 72 J without considering the angle. When teaching this concept, emphasize drawing free body diagrams to visualize the force components. Have students practice identifying which angle to use in the cosine function - it's always the angle between the force vector and displacement vector.
6. A block slides down a frictionless inclined plane of height $h$ and angle $\theta$. At the bottom, it encounters a rough horizontal surface with coefficient of kinetic friction $\mu_k$ and slides a distance $d$ before coming to rest. What is the total work done by friction on the block during its entire motion?
   Answer: $-\mu_k mg d$ Friction only acts on the rough horizontal surface, not on the frictionless incline. The work done by friction is $W_f = -f_k \cdot d = -\mu_k mg d$. Choice B represents the total energy dissipated but doesn't specify work by friction alone. Choice C incorrectly applies the incline angle to the horizontal motion. Choice D incorrectly assumes friction acts along the entire path including the incline.
7. A $4.0\,\text{kg}$ crate is pulled along a frictionless floor by a constant force $F=50\,\text{N}$ directed $60^\circ$ above the horizontal. The crate moves $d=3.0\,\text{m}$ horizontally. Based on the scenario described above, how much work is done by the force in pulling the object? Use $W=Fd\cos(\theta)$ with $\theta=60^\circ$.
   Answer: $75\,\text{J}$ This question tests understanding of work when force is applied at an angle to the displacement. Work equals the product of force and displacement in the direction of motion, W = Fd cos(θ). In this scenario, a 50 N force is applied at 60° above the horizontal while the crate moves 3.0 m horizontally. Choice B is correct because it properly applies the formula: W = (50 N)(3.0 m)cos(60°) = 150 × 0.5 = 75 J. Choice A incorrectly calculates 50 × 3 = 150 J without considering the angle. When teaching this concept, use component analysis to show that only the horizontal component of force (F cos 60°) contributes to work. Practice problems should vary the angle to reinforce that work depends on the force component parallel to displacement.
8. A spring gun fires a ball of mass $m$ vertically upward. The spring has spring constant $k$ and is initially compressed by distance $x_0$. Air resistance does work $W_r$ (where $W_r < 0$) on the ball during its upward flight to maximum height $h$. What is the work done by gravity on the ball from launch to maximum height?
   Answer: $-mgh$ When analyzing work problems in mechanics, focus on the definition: work is the energy transferred to or from an object by a force acting through a displacement. The sign of work depends on whether the force aids or opposes the motion. For the work done by gravity on the ball during its upward flight, consider that gravity always acts downward with magnitude $mg$, while the ball moves upward through displacement $h$. Since the gravitational force opposes the upward motion, gravity does negative work on the ball. Using $W = \vec{F} \cdot \vec{d}$, the work done by gravity equals $W_g = mg \cdot h \cdot \cos(180°) = -mgh$. Option A ($mgh$) incorrectly gives positive work, suggesting gravity aids the upward motion, which contradicts the fact that gravity opposes upward movement. Option B ($\frac{1}{2}kx_0^2 + W_r$) represents the total energy analysis but confuses work done by gravity with the overall energy balance. This expression would be relevant for finding the ball's kinetic energy at launch, not gravity's work. Option C ($\frac{1}{2}kx_0^2 - mgh$) also misapplies energy conservation principles and doesn't isolate the work done specifically by gravity. Remember that work done by individual forces depends only on that force and the displacement, regardless of other forces present. Don't let the presence of air resistance or spring energy confuse you—when asked about work done by gravity alone, focus solely on gravity's force and the displacement.
9. A particle moves along a path such that its position is given by $\vec{r}(t) = (t^2 \hat{i} + 2t \hat{j})$ m. A constant force $\vec{F} = (5 \hat{i} - 3 \hat{j})$ N acts on the particle. What is the instantaneous rate at which work is being done on the particle at $t=1$ s?
   Answer: 4 W The instantaneous rate at which work is done is power, given by $P = \vec{F} \cdot \vec{v}$. First, find the velocity vector by taking the derivative of the position vector: $\vec{v}(t) = \frac{d\vec{r}}{dt} = (2t \hat{i} + 2 \hat{j})$ m/s. At $t=1$ s, the velocity is $\vec{v}(1) = (2 \hat{i} + 2 \hat{j})$ m/s. Now, calculate the power: $P = (5 \hat{i} - 3 \hat{j}) \cdot (2 \hat{i} + 2 \hat{j}) = (5)(2) + (-3)(2) = 10 - 6 = 4$ W.
10. A variable force $F(x) = ax + b$ acts on a particle moving along the x-axis from $x = 0$ to $x = L$. If the work done by this force is $W$, and the force at $x = 0$ is $F_0$ while the force at $x = L$ is $F_L$, which expression correctly relates these quantities?
    Answer: $W = \frac{(F_0 + F_L)L}{2}$ For $F(x) = ax + b$, we have $F_0 = b$ and $F_L = aL + b$. The work is $W = \int_0^L (ax + b)dx = [\frac{ax^2}{2} + bx]_0^L = \frac{aL^2}{2} + bL = \frac{aL^2}{2} + F_0 L = \frac{(aL + b) + b}{2} \cdot L = \frac{F_L + F_0}{2} \cdot L$. This is the average force times displacement. Choice A uses only initial force. Choice B uses only final force. Choice D incorrectly uses the difference rather than average.
11. A satellite of mass $m$ orbits the Earth in a perfect circle of radius $R$. What is the work done by the gravitational force on the satellite as it completes half of an orbit?
    Answer: Zero The gravitational force on the satellite is always directed towards the center of the Earth. The satellite's velocity is always tangent to the circular path. Therefore, the force vector is always perpendicular to the displacement vector at every point in the orbit. Since work requires a component of force parallel to displacement, the work done by gravity on the satellite is zero at all times, including over half an orbit.
12. An object of mass 2 kg moves along a straight line. Its speed changes from 4 m/s to 6 m/s. What is the total work done on the object during this change?
    Answer: 20 J According to the work-energy theorem, the total work done on an object equals its change in kinetic energy: $W_{net} = \Delta K = K_f - K_i$. The initial kinetic energy is $K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(2 \text{ kg})(4 \text{ m/s})^2 = 16 \text{ J}$. The final kinetic energy is $K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(2 \text{ kg})(6 \text{ m/s})^2 = 36 \text{ J}$. The total work done is $W_{net} = 36 \text{ J} - 16 \text{ J} = 20 \text{ J}$.
13. An object is moved from point P to point Q along two different paths, Path 1 and Path 2. The gravitational force and a kinetic friction force are the only forces doing work on the object. Path 2 is longer than Path 1. Let $W_{g1}$ and $W_{f1}$ be the work done by gravity and friction along Path 1, and $W_{g2}$ and $W_{f2}$ be the work done along Path 2. Which statement is correct?
    Answer: $W_{g1} = W_{g2}$ and $|W_{f1}| < |W_{f2}|$ Gravity is a conservative force, so the work it does depends only on the change in vertical position, not on the path taken. Since the start and end points (P and Q) are the same, the change in vertical position is the same for both paths, thus $W_{g1} = W_{g2}$. Friction is a nonconservative force, and the work it does is path-dependent. The work done by friction is $W_f = -f_k \times (\text{path length})$. Since Path 2 is longer, the magnitude of the negative work done by friction will be greater along Path 2, meaning $|W_{f1}| < |W_{f2}|$.
14. A box of mass $M$ is lifted vertically by a cable with an upward acceleration of $a = \frac{g}{4}$. What is the work done by the tension in the cable when the box has been lifted a height $h$?
    Answer: $\frac{5}{4}Mgh$ First, find the tension $T$ using Newton's second law. The net force is $F_{net} = T - Mg = Ma$. Solving for tension, $T = Mg + Ma = M(g+a)$. Substituting $a = g/4$, we get $T = M(g + g/4) = \frac{5}{4}Mg$. The work done by the tension is $W_T = T \cdot h$, since tension and displacement are in the same direction. Therefore, $W_T = \frac{5}{4}Mgh$.
15. A particle is moving to the right. A net force is exerted on the particle. If the net work done on the particle is negative, which of the following must be true?
    Answer: The particle's speed decreases. The work-energy theorem states that the net work done on a particle is equal to the change in its kinetic energy ($W_{net} = \Delta K$). Since kinetic energy is $K = \frac{1}{2}mv^2$, it is always non-negative. If the net work is negative, the change in kinetic energy is negative ($\Delta K < 0$), which means the final kinetic energy is less than the initial kinetic energy. This implies that the particle's speed must decrease.
16. A block slides at a constant velocity across a horizontal, frictionless surface. A constant vertical force of magnitude F is then applied to the block. How much work is done by this new vertical force as the block continues to travel a horizontal distance $d$?
    Answer: Zero Work is done only when there is a component of force in the direction of displacement. The applied force is vertical, while the block's displacement is horizontal. Since the force and displacement vectors are perpendicular to each other, the dot product of the two is zero, and no work is done by the vertical force.
17. A block slides down a curved frictionless track from height $h_1$ to height $h_2$, where $h_1 > h_2$. During this motion, the work done by the track's normal force on the block is:
    Answer: $0$ When analyzing work done by forces, you need to remember that work equals force times displacement in the direction of the force: $W = \vec{F} \cdot \vec{d}$. The key insight here is understanding the direction of the normal force throughout the motion. The normal force always acts perpendicular to the surface at every point along the track. As the block slides down the curved path, the normal force continuously points toward the center of curvature, while the block's displacement follows the tangent to the curve. Since perpendicular vectors have a dot product of zero, the normal force does no work on the block at any instant during the motion. Looking at the incorrect answers: Choice A, $mg(h_1 - h_2)$, represents the work done by gravity (positive because gravity acts downward and the displacement has a downward component). Choice C, $-mg(h_1 - h_2)$, would be the work done against gravity, but this doesn't relate to the normal force. Choice B suggests the track's shape matters, but regardless of whether the track is a gentle slope or has sharp curves, the normal force remains perpendicular to the motion everywhere. You can verify this using energy conservation: the block's kinetic energy increases by $mg(h_1 - h_2)$ due to gravity alone. If the normal force did work, energy wouldn't be conserved in this frictionless system. Strategy tip: When asked about work done by normal forces on curved paths, remember they're always perpendicular to motion. Only forces with components parallel to displacement can do work.
18. A constant force $\vec{F} = (3 \hat{i} + 4 \hat{j})$ N acts on a particle that undergoes a displacement $\Delta\vec{r} = (2 \hat{i} - 2 \hat{j})$ m. What is the work done by the force?
    Answer: -2 J Work is the dot product of the force and displacement vectors: $W = \vec{F} \cdot \Delta\vec{r}$. The dot product is calculated by multiplying corresponding components and summing the results: $W = (F_x)(\Delta x) + (F_y)(\Delta y) = (3 \text{ N})(2 \text{ m}) + (4 \text{ N})(-2 \text{ m}) = 6 \text{ J} - 8 \text{ J} = -2 \text{ J}$.
19. An elevator motor lifts a $650\,\text{kg}$ load vertically upward through a height $h=12\,\text{m}$ at constant speed. Take $g=9.8\,\text{m/s}^2$. The motor's cable tension is upward and approximately equal to the weight, so the force is parallel to the displacement ($\theta=0^\circ$). Based on the scenario described above, determine the work required to lift the mass to the specified height using $W=Fd\cos(\theta)$.
    Answer: $7.6\times10^4\,\text{J}$ This question tests understanding of work done against gravity when lifting an object vertically. Work is the product of force and displacement in the direction of the force, W = Fd cos(θ). In this scenario, the motor must apply an upward force equal to the weight (mg = 650 kg × 9.8 m/s² = 6370 N) to lift at constant speed, moving the load 12 m upward. Choice B is correct because the force and displacement are both upward (θ = 0°), so W = (6370 N)(12 m)(1) = 76,440 J ≈ 7.6×10⁴ J. Choice A incorrectly assigns a negative sign, which would apply if the force opposed the motion. When teaching, emphasize that work is positive when force and displacement are in the same direction. Have students identify force direction relative to motion to determine the sign of work.
20. A 5 kg block is pushed up a 30° incline of length 10 m by a constant force of 60 N parallel to the incline. The coefficient of kinetic friction between the block and the incline is 0.20. What is the net work done on the block? (Use $g=10$ m/s² and $\cos(30^\circ) \approx 0.87$).
    Answer: 263 J The net work is the sum of the work done by the applied force, gravity, and friction. Work by applied force: $W_F = (60 \text{ N})(10 \text{ m}) = 600 \text{ J}$. Work by gravity: $W_g = -mgh = -mgd\sin\theta = -(5)(10)(10)\sin(30^\circ) = -250 \text{ J}$. Normal force: $N = mg\cos\theta = (5)(10)\cos(30^\circ) \approx 43.5 \text{ N}$. Work by friction: $W_f = -f_k d = -\mu_k N d = -(0.20)(43.5)(10) = -87 \text{ J}$. Net work: $W_{net} = 600 - 250 - 87 = 263 \text{ J}$.

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