📝 Work done by force vector (27 MCQs)

📖 Explanation: Work is defined as the line integral $W = \int_C \vec{F} \cdot d\vec{r}$. Reversing the orientation of the curve $C$ negates the differential displacement vector $d\vec{r}$, which consequently negates the entire integral. This demonstrates that work is not merely a function of endpoints but depends critically on the direction of traversal for non-conservative fields.

📖 Explanation: The definition of work requires the dot product $\vec{F} \cdot \vec{d}$, not the product of magnitudes. While $|\vec{F}||\vec{d}|$ equals work only when vectors are perfectly aligned, the general formula is $|\vec{F}||\vec{d}|\cos\theta$. In this specific case they happen to be parallel, but the *reasoning* cited is flawed because it ignores the angular relationship inherent in the definition of mechanical work.

📖 Explanation: Path independence requires the field to be conservative, meaning $\nabla \times \vec{F} = \vec{0}$. Checking the curl components reveals that $\frac{\partial}{\partial x}(y) - \frac{\partial}{\partial z}(x^2+z) = 0 - 1 = -1 \neq 0$. Since the curl is non-zero, the field is non-conservative, and work depends on the specific trajectory taken, regardless of the algebraic form of the components.

📖 Explanation: Gravity is a conservative force depending solely on vertical displacement $\Delta z$. The parameterization shows $z(t) = t$, so $\Delta z = 2\pi - 0 = 2\pi$. The work is $\vec{F} \cdot \Delta \vec{r} = -mg(2\pi)$. Horizontal circular motion contributes nothing since $\vec{F}_{gravity} \perp d\vec{r}_{horizontal}$, simplifying the line integral to a simple potential energy difference calculation.

📖 Explanation: Work is the signed area under the $F-x$ curve relative to the direction of motion. Moving from $b$ to $a$ reverses the integration limits: $\int_b^a F(x)dx = -\int_a^b F(x)dx$. Since the original area from $a$ to $b$ is negative (below axis), the negation makes the work positive. This highlights that physical work sign depends on both force direction and displacement orientation.

📖 Explanation: A vector field is conservative if and only if the line integral vanishes for *every* closed path in the domain. Finding zero work for a single specific loop is insufficient; the field could be non-conservative elsewhere or have cancellation effects unique to that geometry. Rigorous verification requires checking $\nabla \times \vec{F} = \vec{0}$ throughout the simply-connected domain or testing arbitrary paths.

📖 Explanation: Observing that $\vec{F} = \nabla f$ where $f(x,y,z)=xyz$ immediately identifies the field as conservative. Work becomes $f(1,1,1) - f(0,0,0) = 1$. Direct integration would require parameterizing a path, computing derivatives, substituting into the dot product, and integrating. Recognizing the gradient structure transforms a calculus problem into simple arithmetic, demonstrating the power of identifying conservative fields in three-dimensional space.

📖 Explanation: The line integral formula is \int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) dt. The peer's setup uses $\vec{F}(t)$ directly and dots with position $\vec{r}(t)$ instead of velocity \vec{r}'(t). Force must be composed with the path function, and work accumulates along displacement (velocity), not position. This error confuses the parametric representation of the field with the kinematic quantity required for work integration.

📖 Explanation: For conservative fields, work depends exclusively on endpoints, making it identical for all paths connecting A and B. Observing different work values (positive vs negative) for distinct paths between identical endpoints violates path independence. This is the definitive operational test for non-conservativeness, regardless of field complexity, divergence properties, or path lengths, as it directly contradicts the fundamental theorem of line integrals.

📖 Explanation: Notice $\vec{F} = \nabla(\frac{x^3+y^3+z^3}{3})$, confirming conservativeness. Work equals potential difference: $[\frac{(-1)^3+0+(-1)^3}{3}] - [\frac{1^3+0+1^3}{3}] = -\frac{2}{3} - \frac{2}{3} = -\frac{4}{3}$. Wait—rechecking: actually $\vec{F}=\langle x^2,y^2,z^2\rangle$ IS conservative with potential $f=\frac{x^3+y^3+z^3}{3}$. Evaluating correctly gives $-4/3$. However, option A suggests zero; let me reconsider if there's symmetry. Actually the correct answer should be C based on calculation, but this tests whether students blindly assume symmetry yields zero versus performing rigorous potential evaluation on asymmetric endpoints.

📖 Explanation: This rotational field has $\nabla \times \vec{F} = \langle 0,0,2 \rangle \neq \vec{0}$, confirming non-conservativeness. On a circle, $\vec{F}$ is everywhere tangent and parallel to $d\vec{r}$, so $\vec{F} \cdot d\vec{r} > 0$ continuously. Unlike conservative fields where return-to-start guarantees zero work, non-conservative fields can accumulate work over closed loops when the field aligns persistently with the trajectory direction.

📖 Explanation: Verifying $\nabla \times \vec{F} = \vec{0}$ confirms conservativeness. Integrating components yields potential $f = e^x \cos y + z^2/2$. Work becomes $f(\ln 2, \pi/2, 3) - f(0,0,0) = (2 \cdot 0 + 9/2) - (1 \cdot 1 + 0) = 3.5$. This avoids messy trigonometric-exponential integration along arbitrary paths, showcasing how recognizing exact differentials transforms complex line integrals into elementary evaluations.

📖 Explanation: Actually both setups need scrutiny. Correct computation: $\vec{F}(\vec{r}(t)) = \langle 2t,3t,4t \rangle$, \vec{r}'(t)=\langle 1,1,1 \rangle, dot product = $9t$, integral = $9/2 = 4.5$. The first student got the right numerical answer but their written expression omitted showing \vec{r}'(t), suggesting conceptual confusion even if arithmetic was accidentally correct. True understanding requires explicit inclusion of velocity in the integrand formulation.

📖 Explanation: Net work equals the algebraic sum of signed areas: positive for regions above axis (force and displacement aligned), negative below (opposed). Triangle area = $0.5 \times 4 \times h$, rectangle = 3 \times |h'|. Subtracting reflects energy transfer directions. Simply adding magnitudes ignores physics of opposing forces, while ignoring negative regions violates conservation principles. Graphical interpretation requires careful attention to sign, not just geometric area measurement.

📖 Explanation: Mechanical work requires force component along displacement. Since $\vec{F}_B \perp \vec{v}$ always, $\vec{F}_B \cdot d\vec{r} = \vec{F}_B \cdot \vec{v}dt = 0$. Zero instantaneous work means no energy transfer via magnetic forces, making potential energy undefined (potentials describe stored energy convertible to work). This isn't about field uniformity or curl properties but about the fundamental orthogonality preventing work accumulation, distinguishing magnetic from electric forces.

📖 Explanation: Decompose: horizontal part $\langle y,-x,0 \rangle \cdot \langle \cos t, -\sin t, 0 \rangle = y\cos t + x\sin t = \cos^2 t + \sin^2 t = 1$. Vertical: $1 \cdot 1 = 1$. Total integrand = 2, integrated over $2\pi$ gives $4\pi$. Wait—recalculating horizontal: $\vec{F}_h \cdot \vec{v}_h = \cos t \cdot \cos t + (-\sin t)(-\sin t) = \cos^2 t + \sin^2 t = 1$. Plus vertical 1 gives 2. Integral is $4\pi$. So answer should be C. This tests careful decomposition versus intuitive guessing about cancellation.

📖 Explanation: Omitting chain rule during substitution typically means failing to compose $\vec{F}$ with $\vec{r}(t)$ properly or mishandling derivatives. If $\vec{F}$ isn't correctly expressed as function of t, the dot product with \vec{r}'(t) produces terms with wrong dimensional structure. Work must have energy units (N·m); getting N·s indicates temporal rather than spatial integration, revealing failure to properly transform the vector field into the parameter domain before integration.

📖 Explanation: Radial motion: $\vec{F} \parallel d\vec{r}$, work = $\int_0^R r \cdot dr = R^2/2 > 0$. Tangential motion on sphere: $\vec{F}$ is radial while $d\vec{r}$ is tangential, so $\vec{F} \perp d\vec{r}$ everywhere, yielding zero work. This illustrates how work depends on relative orientation, not just force magnitude or path length, and how central forces do no work during perpendicular displacements despite being nonzero.

📖 Explanation: Upward journey: $W_{up} = \int_0^h kz \, dz = kh^2/2$. Downward: $W_{down} = \int_h^0 kz \, dz = -kh^2/2$. Sum = 0 per cycle. Despite force being position-dependent and nonzero, the conservative nature ensures cyclic work vanishes. This contrasts with dissipative forces where cyclic work is negative, highlighting how conservativeness manifests in periodic motion regardless of force complexity.

📖 Explanation: Green's theorem $\oint_C P dx + Q dy = \iint_D (Q_x - P_y) dA$ requires D to be simply connected and field defined/smooth throughout. If field has singularities inside D (e.g., $\vec{F} = \langle -y/(x^2+y^2), x/(x^2+y^2) \rangle$ at origin), theorem fails even if curl appears zero algebraically. Students frequently apply it to punctured domains, obtaining incorrect zero work when actual circulation is nonzero, missing topological constraints.

📖 Explanation: Check if $\vec{F} = \nabla f$. If $f_x = 3t^4$ etc., but F is given in t, not xyz. Reparameterize F in spatial terms: since $x=t^2, y=t^3, z=t$, we get $t=z$, so $\vec{F} = \langle 3z^4, 2z^2, 1 \rangle$. Now check if conservative: $\partial F_z/\partial y = 0$, $\partial F_y/\partial z = 4z \neq 0$. Not conservative. So simplification fails; direct integration needed. This tests premature optimization attempts versus rigorous verification.

📖 Explanation: For $\vec{F} = \langle -y,x,0 \rangle$, work density $\vec{F} \cdot d\vec{r} = -y dx + x dy$. On semicircle radius R: parametrization gives constant positive integrand $R^2 d\theta$, total $\pi R^2$. Straight line through origin: symmetric cancellation yields less work. Curved path maintains favorable force-displacement alignment, accumulating more work. This demonstrates how geometry interacts with field structure in non-conservative systems, beyond mere endpoint dependence.

📖 Explanation: If $\vec{F} \cdot \vec{v} = 0$ holds for ALL t in the interval, then $W = \int \vec{F} \cdot \vec{v} \, dt = 0$. The student's logic is actually correct IF the premise is universally true. However, if the condition holds only at isolated instants or intermittently, work accumulates during other periods. The key distinction is between pointwise and continuous satisfaction of the orthogonality condition, testing precise mathematical reasoning about integration of instantaneous quantities.

📖 Explanation: Quick curl check: all mixed partials match, confirming $\vec{F} = \nabla(xy+xz+yz)$. Work = $f(1,2,3)-f(0,0,0) = (2+3+6)-0 = 11$. This bypasses parameterization entirely. Students wasting time on line integrals miss the structural insight. Efficiency comes from pattern recognition in symmetric expressions, transforming multivariable calculus into algebraic evaluation, emphasizing conceptual fluency over computational brute force.

📖 Explanation: If path 3 is genuinely a different route from B to A, its work needn't relate to paths 1 or 2. But if path 3 is supposed to be the reverse of path 1 or 2, then $W_{BA} = -W_{AB}$ MUST hold by line integral properties regardless of conservativeness. Getting -12J instead of -10J or -15J indicates either measurement error or misidentification of the return path. This tests understanding that reversal symmetry is universal, unlike path independence.

📖 Explanation: Drag opposes velocity by construction: $\vec{F}_d = -k|\vec{v}|\hat{v} = -k|\vec{v}|\vec{v}/|\vec{v}|$. Dot product: $\vec{F}_d \cdot \vec{v} = -k|\vec{v}|(\vec{v} \cdot \vec{v})/|\vec{v}| = -k|\vec{v}|^2$. Wait—actually $\vec{F}_d = -k|\vec{v}|\vec{v}$ gives $\vec{F}_d \cdot \vec{v} = -k|\vec{v}|(\vec{v}\cdot\vec{v}) = -k|\vec{v}|^3$. Either way, strictly negative for nonzero velocity. This guarantees energy dissipation, modeling irreversibility fundamentally through the mathematical structure of the force law itself.

📖 Explanation: Algebraically $\nabla \times \vec{F} = \vec{0}$ where defined, but domain excludes z-axis, creating a hole. The potential $\theta = \arctan(y/x)$ is multi-valued, increasing by $2\pi$ per revolution. Work = $\oint d\theta = 2\pi \neq 0$. This classic example shows curl-free ≠ conservative when topology prevents global potential definition. Students must recognize that vector calculus theorems have topological preconditions beyond local differential conditions, linking analysis to geometry profoundly.