Work-Energy Principle - Statics and Dynamics

1. A student attempts to apply the work–energy principle to find the final speed $v_2$ by computing friction work as $W_f = -(6 + 2v_1)(3) = -(6+16)(3) = -66 \text{ J}$, then solving $\tfrac{1}{2}(4)(8)^2 - 66 = \tfrac{1}{2}(4)v_2^2$. Which statement correctly identifies the error and its consequence?
   Answer: The student's error is treating the velocity-dependent friction force as constant at its initial value; since the friction force varies with speed, the work must be computed as $W_f = -\int_0^3 f\,dx$, which requires expressing $f$ as a function of position (via the equation of motion) rather than directly integrating over $x$, making the work–energy approach non-trivial for this force law. Whenever you encounter a velocity-dependent force in a work–energy problem, your first instinct should be to ask: can I treat this force as constant? If the force changes as the particle moves, you cannot simply plug in one speed value and multiply by distance. The work–energy principle itself is perfectly valid here — the issue is how the work is computed. The correct friction work is $W_f = -\int_0^3 f\, dx$, but since $f = 6 + 2v$ and $v$ changes continuously along the path, you cannot integrate directly over $x$ without first knowing $v(x)$. To find $v(x)$, you must solve the equation of motion $m\frac{dv}{dt} = -f(v)$, which typically involves converting to $v\frac{dv}{dx}$ form and integrating — a non-trivial step. This is exactly what answer B captures: the student's error is freezing the force at its initial value, and the remedy requires relating the force law to position through the dynamics, making the work–energy approach more involved than it first appears. A is tempting but wrong — using an average speed is an ad hoc approximation with no rigorous justification, and iterating to convergence is not a standard method for this type of problem. C describes a sign error that doesn't actually exist. The student correctly used a negative sign for friction work; adding 66 J would increase kinetic energy, which is physically absurd for a friction force. D is a fundamental misconception. The work–energy principle applies to all forces, conservative or not — velocity-dependent forces are no exception. Study tip: When a force depends on velocity, always check whether you can express it as a function of position before applying $W = \int F\,dx$. If you can't, the integral isn't straightforward.
2. A particle of mass $m$ is released from rest at the top of a frictionless hemispherical bowl of radius $R$. Using the work–energy principle alone (without Newton's second law), which of the following quantities can be determined at the point where the particle has descended a height $h$ below the rim?
   Answer: Only the speed of the particle, because the work–energy principle is a scalar equation relating kinetic energy changes to work done and yields speed directly; the normal force does no work and therefore does not appear in — and cannot be extracted from — the work–energy equation alone. Whenever a problem asks what the work–energy principle "alone" can determine, your first move should be to write down the equation and see exactly which unknowns appear in it. The work–energy theorem states that the net work done on a particle equals its change in kinetic energy: $W_{net} = \Delta KE$. For a particle sliding down a frictionless bowl, only gravity does work (the normal force is always perpendicular to the motion, so it contributes zero work). Starting from rest at the rim and descending height $h$, the equation becomes: $mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh}$ Speed pops out directly — no other law needed. This confirms B is correct. Now notice what's missing from that equation: the normal force $N$ never appears, precisely because it does no work. To find $N$, you must apply Newton's second law in the radial direction, setting the net inward force equal to the centripetal acceleration: $N - mg\cos\theta = \frac{mv^2}{R}$. You can substitute $v^2 = 2gh$ here, but the point is that Newton's second law — not the work–energy principle — is required to extract $N$. A is wrong because it claims work–energy alone yields the normal force, which it cannot. C has the logic exactly backwards: it's the speed, not the normal force, that comes directly from work–energy. D contains a fundamental misconception — the work–energy theorem is completely general and applies to any path, curved or straight. Study tip: Whenever a problem involves a constraint force (normal, tension), remember that constraint forces are typically perpendicular to motion, do no work, and are invisible to the work–energy equation — you'll always need Newton's second law to find them.
3. Using the work–energy principle applied from start to the point of maximum spring compression, which expression correctly represents the work–energy equation? (Take $g = 9.81 \text{ m/s}^2$, and let $d$ be the length of the rough horizontal section.)
   Answer: $0 + mg(3) - \mu_k mg(d + 0.2) - \tfrac{1}{2}k(0.2)^2 = 0$, because all work terms — gravity, friction over the full contact distance including spring compression, and spring elastic potential — must sum to zero when the final kinetic energy is zero. When applying the work–energy principle to a system with multiple forces, your equation must account for every force doing work over every portion of the path — from start to finish. Set it up as: $\Delta KE = \sum W_{all forces}$. Since the particle starts and ends at rest, $\Delta KE = 0$, so all work terms must sum to zero. Here's the key insight for this problem: friction acts over the entire horizontal sliding distance, which includes both the flat section $d$ and the additional 0.2 m the particle travels while compressing the spring. The spring also does negative work on the particle equal to $\tfrac{1}{2}k(0.2)^2$. This gives: $0 + mg(3) - \mu_k mg(d + 0.2) - \tfrac{1}{2}k(0.2)^2 = 0$, confirming that A is correct. B is tempting but wrong — it omits friction over the spring-compression distance. The surface is still horizontal and rough during those 0.2 m, so friction doesn't suddenly disappear when the spring is contacted. C makes two errors at once: it applies friction correctly over $d + 0.2$, but then treats the spring's elastic energy as a positive kinetic energy gain rather than negative work done on the particle. The spring opposes motion, so it does negative work. D ignores the spring work entirely, claiming elastic energy "cancels itself out." This is a phantom cancellation — the spring's negative work absolutely belongs in the equation. Your strategy: always trace the particle's full path and ask, "Is this force active here?" for each segment. Friction follows the particle everywhere it slides on a rough surface — including during spring compression.
4. A particle of mass $m$ travels along a curved frictionless path in the vertical plane. At position 1 it has speed $v_1$ and at position 2 (which is higher by $\Delta h$) it has speed $v_2$. A non-conservative force $\mathbf{F}_{nc}$ acts on the particle between positions 1 and 2. A student writes the work–energy equation as $\tfrac{1}{2}mv_1^2 + W_{nc} = \tfrac{1}{2}mv_2^2 + mg\Delta h$. Which statement best characterizes this equation?
   Answer: The equation is correct as written; it is a valid form of the work–energy principle where gravitational potential energy change $mg\Delta h$ appears on the right side to account for the energy stored against gravity, and $W_{nc}$ represents all non-conservative work inputs. Whenever you see a work–energy problems involving gravity, the key question is whether gravity is treated as a work term or a potential energy term — both approaches are valid, but you must apply them consistently. The work–energy principle in its most general form states that the net work done on a particle equals its change in kinetic energy: $W_{net} = \Delta T$. Gravity's work over a rise of $\Delta h$ is $W_g = -mg\Delta h$. If you move that term to the right side, the equation becomes $T_1 + W_{nc} = T_2 + mg\Delta h$, which is exactly what the student wrote. This is simply algebraic rearrangement — the physics is identical. The right-hand $mg\Delta h$ represents the gain in gravitational potential energy, and $W_{nc}$ captures all non-conservative work. The equation remains fully valid whether or not $W_{nc} = 0$. So B is correct. A is wrong because it claims moving gravity to the right side restricts the equation to conservative-only situations. That's false — algebraic rearrangement never changes the physical content. The equation handles $W_{nc} \neq 0$ perfectly. C is wrong because it insists on a rigid "standard form" where all work appears on the left. There is no such restriction; physics equations can be rearranged freely as long as the algebra is correct. Equivalent forms are equally valid. D is wrong because it confuses two entirely separate theorems. Impulse ($\int \mathbf{F}\, dt$) governs momentum changes; work ($\int \mathbf{F} \cdot d\mathbf{r}$) governs energy changes. Non-conservative forces appear naturally in the work–energy theorem — no substitution is needed. Study tip: When you see $mg\Delta h$ on either side of a work–energy equation, ask yourself whether it's being treated as work or potential energy — as long as you're consistent and the algebra checks out, the form is valid.
5. After the system has moved a distance $d$ (B descends by $d$, A moves horizontally by $d$), applying the work–energy principle to the system as a whole yields which correct equation for the common speed $v$?
   Answer: $\tfrac{1}{2}(3m)v^2 = mgd$, because the only external force doing work on the system is gravity acting on B (the table's normal force and the cord tension are internal or do no work), and both particles share the same speed due to the inextensible cord. When applying the work–energy theorem to a system of particles, your job is to identify which external forces do work on the system, then set that total work equal to the change in kinetic energy of all particles combined. Here, the system consists of both A and B. The external forces acting are: gravity on B ($mg$ downward), gravity on A (balanced by the table's normal force), the normal force itself, and the cord tension. The normal force does no work (perpendicular to motion). The cord tension is internal to the system — it acts between the two particles and its work contributions cancel exactly. Gravity on A does no work because A moves horizontally. That leaves only gravity on B doing net work: $W_{net} = mgd$. Since both particles move at the same speed $v$ (inextensible cord), the total kinetic energy is $\tfrac{1}{2}(2m)v^2 + \tfrac{1}{2}(m)v^2 = \tfrac{1}{2}(3m)v^2$. Setting these equal gives $\tfrac{1}{2}(3m)v^2 = mgd$, confirming C. A is wrong because you cannot ignore A's kinetic energy just because it isn't directly driven by gravity — all moving masses contribute to the system's kinetic energy. B is wrong because A moves horizontally, not vertically, so gravity does zero work on A; there is no "$-2mgd$" term. D is the most tempting trap: it treats the cord tension as if it were an external force that removes energy from the system. But tension is internal — it transfers energy between particles without adding or removing energy from the system as a whole. Study tip: Whenever you apply the work–energy theorem to a system, immediately classify every force as internal or external, then ask whether each external force is parallel to displacement. Internal forces never appear in the net work term.
6. What is the speed of the crate at the end of the 5 m travel? (Take $g = 9.81 \text{ m/s}^2$.)
   Answer: $v \approx 7.05 \text{ m/s}$, from $\tfrac{1}{2}mv^2 = T(5) - mg\sin30°(5) - \mu_k mg\cos30°(5)$, correctly accounting for tension work, the gravity component along the incline, and friction over 5 m. When a problem involves a moving object on an incline with friction, your go-to framework is the work-energy theorem: the net work done on the object equals its change in kinetic energy. Starting from rest, that means $\tfrac{1}{2}mv^2 = W_{net}$. The key skill is correctly identifying every force doing work — and computing each work term using only the component of force along the direction of motion. For this crate moving up the incline, three forces do work over the 5 m displacement: the rope tension (positive work), gravity's component along the incline (negative work), and kinetic friction (negative work). The normal force does no work since it's perpendicular to motion. The correct equation is $\tfrac{1}{2}mv^2 = T(5) - mg\sin30°(5) - \mu_k mg\cos30°(5)$. Plugging in: tension work = $120(5) = 600\ \text{J}$; gravity work = $10(9.81)(0.5)(5) = 245.25\ \text{J}$; friction work = $0.25(10)(9.81)(0.866)(5) = 106.5\ \text{J}$. So $\tfrac{1}{2}(10)v^2 = 600 - 245.25 - 106.5 = 248.25\ \text{J}$, giving $v \approx 7.05\ \text{m/s}$. This confirms D is correct. A is close but uses the full weight $mg$ instead of $mg\cos30°$ for the normal force in the friction term — a classic incline mistake. B neglects friction entirely, which is never valid when $\mu_k > 0$ and the surface is in contact. C compounds two errors: it uses the full weight $mg$ for gravity's work (gravity only acts against the incline component of displacement) and, while it uses $\cos30°$ for friction correctly, the first error produces a wildly low speed. Your strategy: on any inclined-plane problem, immediately draw a free-body diagram and decompose every force into parallel and perpendicular components. Friction uses $mg\cos\theta$; gravity's work uses $mg\sin\theta$. Mixing these up is the single most common error on incline questions.
7. A particle of mass 1.5 kg is launched horizontally with speed $v_0 = 10 \text{ m/s}$ from the edge of a cliff of height $H = 20 \text{ m}$. Air resistance exerts a force whose magnitude is $F_{drag} = 0.3v^2 \text{ N}$, always directed opposite to the velocity vector. The particle lands at the base of the cliff (elevation drop = $H$). Student 1 claims: 'Since air resistance always opposes velocity, the drag work over the entire flight is negative, so the particle's landing speed must be less than the landing speed in vacuum.' Student 2 claims: 'With drag, the trajectory changes — the particle travels a longer curved path, so the drag force acts over a greater path length, potentially increasing the net work done by all forces and raising the landing speed above the vacuum value.' Which student is correct, and why?
   Answer: Student 1 is correct. By the work–energy principle, $T_{landing} = T_0 + W_{gravity} + W_{drag}$. Since gravity is conservative, $W_{gravity} = mgH$ regardless of the path taken — it depends only on the fixed elevation drop $H$. Since $W_{drag} < 0$ always (drag opposes velocity at every instant), the landing kinetic energy with drag is strictly less than without drag. Whenever a problem mixes trajectory changes with energy methods, anchor yourself to the work–energy theorem: $KE_{landing} = KE_0 + W_{gravity} + W_{drag}$. This framework cuts through the confusion immediately. Here's the key insight: gravity is a conservative force, meaning its work depends only on the vertical displacement — not the shape of the path. Since the particle always falls exactly $H = 20\text{ m}$, $W_{gravity} = mgH$ is identical whether drag is present or not. Meanwhile, drag is always directed opposite to velocity, so at every instant the angle between $\vec{F}_{drag}$ and $d\vec{s}$ is exactly 180°, making $dW_{drag} < 0$ at every point along the path. Therefore $W_{drag} < 0$ over the entire flight — guaranteed. Plugging in: $KE_{landing}^{(drag)} = KE_0 + mgH + W_{drag} < KE_0 + mgH = KE_{landing}^{(vacuum)}$. Student 1 is correct, confirming answer C. Answer A is wrong because it claims the particle descends more than $H$. The elevation drop is fixed by the cliff geometry — the particle starts at height $H$ and lands at the base, period. Drag cannot change the endpoint's elevation. Answer B contains a correct observation (drag does act over a longer path, removing even more energy) but mislabels Student 2 as "partially valid." Student 2's core claim — that longer path length could raise landing speed — is simply false, so crediting that reasoning misleads you. Answer D incorrectly treats the height drop as uncertain. Because both launch and landing elevations are fixed, $W_{gravity} = mgH$ is exact, making the comparison fully determinate. Study tip: When you see drag or friction problems, immediately write the work–energy theorem and ask: does the conservative-force work change? Usually it doesn't — then the sign of the non-conservative work decides everything.
8. A 2 kg particle is pulled from rest along a horizontal frictionless surface by a force $F = 10 + 3x \text{ N}$ (where $x$ is displacement in meters) acting in the direction of motion. After traveling 4 m, the particle encounters a 0.5 m drop to a lower frictionless horizontal surface. What is the speed of the particle after it has traveled an additional 2 m along the lower surface? (Take $g = 9.81 \text{ m/s}^2$; assume the applied force ceases at the drop.)
   Answer: $v = \sqrt{\frac{2}{m}\left[\int_0^4(10+3x)\,dx + mg(0.5)\right]} \approx 8.6 \text{ m/s}$, obtained by integrating the variable force over 4 m and adding the gravitational work through the 0.5 m drop; the additional 2 m on the lower frictionless surface contributes no work since no force acts horizontally there. When a problem involves a variable force, gravity, and multiple surface segments, your go-to framework is the work-energy theorem: the net work done on a particle equals its change in kinetic energy. The key discipline is carefully identifying which forces do work and over which distances they act. Start from rest, so $KE_i = 0$. The applied force $F = 10 + 3x$ acts only over the first 4 m on the upper surface. Its work is: $W_F = \int_0^4 (10 + 3x)\,dx = \left[10x + \frac{3x^2}{2}\right]_0^4 = 40 + 24 = 64 \text{ J}$ During the 0.5 m drop, gravity does work $W_g = mgh = 2(9.81)(0.5) = 9.81 \text{ J}$. On the lower surface, no horizontal force acts and the surface is frictionless, so no additional work is done over the final 2 m. Applying the work-energy theorem: $v = \sqrt{\frac{2(64 + 9.81)}{2}} = \sqrt{73.81} \approx 8.6 \text{ m/s}$ This confirms answer A is correct. Answer B incorrectly extends the variable force integral over 6 m total, as if $F$ continues acting on the lower surface — but the problem explicitly states the force ceases at the drop. Answer C neglects gravitational work entirely during the drop. Gravity is a real force doing real work over the 0.5 m descent; it absolutely belongs in the energy accounting. Answer D misapplies an average force over the wrong distance, compounding two errors at once — wrong force duration and invalid averaging of a non-constant force. Study tip: Always map each force to its exact displacement segment before calculating work. Draw a simple sketch labeling where each force starts and stops — this prevents both the "extend everything over total distance" trap (B) and the "forget gravity" trap (C).
9. Using the work–energy principle, determine whether the cord breaks before the ball reaches the top of the circle, and identify the correct reasoning.
   Answer: The cord does not break. The work–energy principle gives $v_{top}^2 = v_{bottom}^2 - 4gL \approx -11.1 \text{ m}^2/\text{s}^2 < 0$, confirming the ball cannot reach the top. The cord is most likely to break at the bottom, where tension is maximum. At the bottom: $T_{bottom} = m\left(\frac{v_{bottom}^2}{L} + g\right) = 0.5\left(\frac{36}{1.2} + 9.81\right) = 0.5(30 + 9.81) = 19.9 \text{ N} < 30 \text{ N}$, so the cord does not break at the bottom. Since this is the maximum tension during the motion, the cord does not break anywhere. When a ball swings in a vertical circle, two tools work together: the work–energy principle tells you the speed at any point, and Newton's second law in the radial direction tells you the tension. The critical insight is identifying where tension is greatest — that's where the cord is most likely to break. Start with energy. The height difference between bottom and top is $2L$, so: $v_{top}^2 = v_{bottom}^2 - 2g(2L) = 36 - 4(9.81)(1.2) \approx -11.1 \; \text{m}^2/\text{s}^2$ A negative value is physically impossible — the ball never reaches the top. Instead, it loses all kinetic energy at some intermediate angle and reverses direction. This means the critical point to check is the bottom, where speed — and therefore centripetal acceleration — is maximum. At the bottom, tension and weight act in opposite directions along the cord, giving: $T_{bottom} = m\!\left(\frac{v_{bottom}^2}{L} + g\right) = 0.5\!\left(\frac{36}{1.2} + 9.81\right) \approx 19.9 \; \text{N} < 30 \; \text{N}$ The cord survives. Because tension only decreases as the ball rises and slows, it never reaches $T_{max}$ anywhere. That's the logic behind answer D. A correctly identifies that the ball can't reach the top but then falsely implies you can't determine the outcome — you absolutely can by checking the bottom. B claims the cord goes slack with zero tension, which would only happen if the ball were on the inside of a loop; here the cord pulls inward throughout the downswing. C uses the wrong height change ($2gL$ instead of $4gL$), incorrectly concluding the ball reaches the top. Study tip: On circular-motion problems, always ask "where is speed highest?" — that's where tension peaks, and that's your critical check point.

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