Answer: $k = 250$ N/m

Determine the displacement: The displacement from the natural length is $x = 0.5 - 0.3 = 0.2$ m. Apply Hooke's Law: From $F = kx$, we get $k = F/x = 50/0.2 = 250$ N/m. Why distractors fail: Option A ($100$ N/m) results from dividing $50$ by $0.5$ (using total length instead of displacement). Option C ($166.7$ N/m) results from $50/0.3$ (dividing by natural length). Option D ($25$ N/m) results from $50 \times 0.5$ or a similar arithmetic error.

Answer: The claim is valid; the net work is zero because the force does positive work over $[0, \pi]$ and an equal amount of negative work over $[\pi, 2\pi]$

Evaluate the integral: $\int_0^{2\pi} \sin(x)\, dx = [-\cos(x)]_0^{2\pi} = (-\cos 2\pi) - (-\cos 0) = -1 + 1 = 0$. Interpret physically: The net work is indeed zero. Over $[0, \pi]$, $\sin(x) > 0$ so the force does positive work. Over $[\pi, 2\pi]$, $\sin(x) < 0$ so the force does negative work. These cancel exactly. Why distractors fail: Option A: net work can be zero or negative; we don't take absolute values unless asked for total energy. Option C: the arithmetic error is fabricated; the integral does equal zero. Option D: work depends on the force and displacement, not on whether the object returns.

Answer: The student is correct: $(h - y)$ is the distance from a slice at height $y$ to the vertex at the top

Clarify the setup: The cone is inverted with vertex at the top (at height $h$). Water is pumped out through the vertex. A slice at height $y$ (from the bottom) must be lifted to $y = h$. Determine the correct distance: The lifting distance from height $y$ to the outlet at height $h$ is $h - y$. The student's integral is correct. Why distractors fail: Option B: $y$ would be the distance from the bottom to the slice, not from the slice to the top. Option C: using a constant $h$ ignores that slices at different heights travel different distances. Option D: the integrals $\int_0^h f(y)(h-y)\,dy$ and $\int_0^h f(y) \cdot y\,dy$ are generally not equal.

Answer: The depth of the horizontal strip below the surface of the fluid

Understanding the hydrostatic pressure formula: Pressure at depth $d$ in a fluid is $P = \rho g d$. In the integral, $d(y)$ gives the depth of a thin horizontal strip below the fluid surface. Why Option C is correct: $d(y)$ converts the coordinate $y$ into a depth measurement, which determines the pressure acting on the strip at that position. Why distractors fail: Option A describes $w(y)$, the width function. Option B describes $\rho$, which is typically constant. Option D is a fixed geometric measurement, not a function of $y$.

Answer: $W = \int_a^b F(x)\, dx$

Recall the definition of work by integration: When force varies with position, work is computed by summing infinitesimal contributions $F(x)\,dx$ over the interval $[a,b]$. Why the correct answer works: $W = \int_a^b F(x)\,dx$ correctly accumulates the variable force over the displacement, giving the total work done. Why distractors fail: Option A confuses work with the net change of a function. Option C applies differentiation rather than integration. Option D uses a constant-force formula $W = F \cdot d$, which only works when force is constant.

Answer: $h - y$

Set up the coordinate system: With $y$ measured from the bottom of the tank, the top rim is at $y = h$. Determine the lifting distance: A slice at height $y$ must be raised to height $h$, so the distance is $h - y$. Why distractors fail: Option A ($y$) gives the distance from the bottom, not the distance remaining to the top. Option C ($h + y$) would mean the slice goes above the tank by $y$. Option D ($h$) treats every slice as if it starts at the very bottom.

Answer: $72$ J

Set up the integral: $W = \int_0^4 (3x^2 + 2)\, dx$ Evaluate: $W = \left[x^3 + 2x\right]_0^4 = (64 + 8) - 0 = 72$ J. Why distractors fail: Option B ($50$) may result from incorrect integration such as $\frac{3x^3}{3}\big|_0^4 = 64$ minus something wrong. Option C ($64$) comes from computing only $x^3\big|_0^4$ and forgetting the $+2x$ term. Option D ($80$) might come from $F(4) \cdot 4 = 50 \cdot ... $ or another arithmetic mistake.

Answer: $W = \int_0^3 9800 \pi (6y - y^2)(3 - y)\, dy$

Determine the cross-sectional radius: For a hemisphere of radius $R = 3$ with center at $(0,3)$ (center of the flat top), the equation is $x^2 + (y-3)^2 = 9$. At height $y$, the cross-sectional radius squared is $r^2 = 9 - (y-3)^2 = 6y - y^2$. Set up the work integral: A slice at height $y$ has volume $\pi(6y - y^2)\,dy$, weight $9800\pi(6y - y^2)\,dy$, and must be lifted $(3 - y)$ to exit at the top. Why distractors fail: Option A uses $9 - y^2$, which is the radius formula for a hemisphere centered at the origin (flat side down with center at bottom). Option C uses $y^2$ for the radius squared, which doesn't match any standard hemisphere orientation. Option D uses the correct expression for a differently oriented hemisphere but has the wrong lifting distance factor $y$ instead of $(3-y)$.

Answer: $V = 2\pi \bar{y} \cdot A$

State Pappus's Theorem: The volume of a solid of revolution is $V = 2\pi d \cdot A$, where $d$ is the distance from the centroid to the axis of revolution and $A$ is the area of the region. Apply to the $x$-axis: When revolving about the $x$-axis, the centroid's distance to the axis is $\bar{y}$. Therefore $V = 2\pi \bar{y} \cdot A$. Why distractors fail: Option A uses $\bar{x}$, which would apply to revolution about the $y$-axis. Option C squares $\bar{y}$, which is not part of Pappus's formula. Option D inverts the relationship.

Answer: $\bar{x} = \frac{1}{A}\int_a^b x\, f(x)\, dx$

Recall the moment integral for $\bar{x}$: The centroid $\bar{x} = \frac{M_y}{A}$ where $M_y = \int_a^b x\, f(x)\, dx$ is the first moment about the $y$-axis. Why Option B is correct: Each vertical strip at position $x$ has area $f(x)\,dx$ and its center is at $x$, so the moment is $x \cdot f(x)\,dx$. Dividing by total area $A$ gives $\bar{x}$. Why distractors fail: Option A is the formula for $\bar{y}$, not $\bar{x}$. Option C gives $A/A = 1$ (just the area divided by itself). Option D omits $f(x)$ entirely and doesn't weight by the region's shape.

Answer: $\bar{y} = 12/5$

Compute the area: $A = \int_{-2}^{2}(4 - x^2)\,dx = 2\int_0^2(4 - x^2)\,dx = 2\left[4x - \frac{x^3}{3}\right]_0^2 = 2(8 - 8/3) = 32/3$. Compute $M_x$: $M_x = \int_{-2}^{2} \frac{1}{2}(4^2 - (x^2)^2)\,dx = \int_{-2}^{2} \frac{1}{2}(16 - x^4)\,dx = 2 \cdot \frac{1}{2}\int_0^2 (16 - x^4)\,dx$. Evaluating: $\int_0^2(16 - x^4)\,dx = [16x - x^5/5]_0^2 = 32 - 32/5 = 128/5$. So $M_x = 128/5$. Find $\bar{y}$: $\bar{y} = M_x / A = (128/5)/(32/3) = (128/5)(3/32) = 384/160 = 12/5$. Why distractors fail: Option A ($8/5$) might arise from an error in the moment integral. Option C ($2$) is the midpoint of $[0,4]$ but doesn't account for the shape. Option D ($16/5$) may result from omitting the $\frac{1}{2}$ in the moment formula for $\bar{y}$.

Answer: The approach is correct and gives the right answer, since $\int_{0.2}^{0.5} 400x\,dx = \frac{400}{2}(0.5^2 - 0.2^2) = 42$ J

Verify by integration: $W = \int_{0.2}^{0.5} 400x\,dx = 400 \cdot \frac{x^2}{2}\Big|_{0.2}^{0.5} = 200(0.25 - 0.04) = 200(0.21) = 42$ J. Assess the student's shortcut: The student used the antiderivative $\frac{k}{2}x^2$ and evaluated at the bounds, which is exactly what the integral yields. The answer $42$ J is correct. Why distractors fail: Option A claims the approach is wrong, but it's algebraically identical to direct integration. Option C uses $F = kx$ with a constant force, ignoring the variable nature. Option D incorrectly claims $1/2$ doesn't belong, but it arises from the power rule antiderivative of $x$.

Answer: $W = \int_0^3 9800 \cdot 10 \cdot \frac{4y}{3} \cdot (3 - y)\, dy$

Set up coordinates and width function: Let $y$ be measured from the vertex (bottom). At $y = 0$, width = 0; at $y = 3$, width = 4. By similar triangles, width $= 4y/3$. Determine lifting distance: The top edge is at $y = 3$. A slice at height $y$ must be lifted $3 - y$ meters. Assemble the integral: Slice volume: $10 \cdot (4y/3)\,dy$. Weight: $9800 \cdot 10 \cdot (4y/3)\,dy$. Work for slice: weight $\times (3 - y)$. This matches Option B. Why distractors fail: Option A uses $y$ instead of $(3 - y)$ for the lifting distance. Option C uses $4(3-y)/3$ for the width, which gives zero width at the top and maximum at the bottom—inverted. Option D uses a constant width of 4, not accounting for the triangular shape.

Answer: Because different slices of liquid must travel different distances to exit the tank

Understand why integration is needed: In a pumping problem, each horizontal slice of liquid is at a different height, so each must be lifted a different distance to reach the top of the tank (or the spout). Why Option C is correct: A slice near the bottom must travel farther than a slice near the top. Since work = force × distance and the distance varies, we must integrate over all slices. Why distractors fail: Option A: for most problems the fluid density $\rho$ is constant. Option B: $g$ is treated as constant. Option D: pressure is relevant for hydrostatic force, not for computing work to lift fluid.

Answer: $w(y) = 4 - 4y/3$

Identify the geometry: The plate has its base of length 4 along the surface ($y = 0$) and tapers to a point at depth $y = 3$. Find the width function: At $y = 0$, width is 4. At $y = 3$, width is 0. The hypotenuse has equation $x/4 + y/3 = 1$, so $x = 4(1 - y/3) = 4 - 4y/3$. The width at depth $y$ is $w(y) = 4 - 4y/3$. Why distractors fail: Option A gives $w(0) = 0$ and $w(3) = 4$, reversing the triangle. Option C uses 3 instead of 4 for the base width. Option D gives $w(1) = 0$ instead of $w(3) = 0$, using the wrong vertex depth.

Answer: The spring constant (stiffness of the spring)

Identify Hooke's Law parameters: In $F(x) = kx$, $k$ is the spring constant that measures how stiff the spring is. A larger $k$ means more force is needed per unit displacement. Why Option B is correct: The spring constant $k$ has units of force per length (e.g., N/m) and directly quantifies the spring's resistance to deformation. Why distractors fail: Option A describes a geometric property, not a force parameter. Option C describes a physical limit, not a proportionality constant. Option D is close but incorrect: the work to stretch by one unit is $\int_0^1 kx\,dx = k/2$, not $k$ itself.

Answer: $980$ J

Set up coordinates: Let $y$ be the distance from the top of the building. A small segment at distance $y$ has mass $0.5\,dy$ and must be lifted $y$ meters. Evaluate the integral: $W = \int_0^{20} 0.5 \cdot 9.8 \cdot y\, dy = 4.9 \cdot \frac{y^2}{2}\Big|_0^{20} = 4.9 \cdot 200 = 980$ J. Why distractors fail: Option A ($490$) results from $4.9 \cdot 100$, perhaps using $y$ instead of $y^2/2$. Option C ($196$) comes from $0.5 \cdot 9.8 \cdot 20 \cdot 2 = 196$, an incorrect formula. Option D ($1960$) doubles the answer, perhaps forgetting the $1/2$ from integration.

Answer: Because each vertical strip extends from $y = 0$ to $y = f(x)$, so its centroid is at $y = \frac{f(x)}{2}$, and multiplying by the strip area $f(x)\,dx$ gives the factor $\frac{1}{2}[f(x)]^2\,dx$

Locate the centroid of a thin strip: A vertical strip at position $x$ extends from $y = 0$ to $y = f(x)$. Its own centroid (midpoint) is at $\tilde{y} = f(x)/2$. Form the moment integral: The moment about the $x$-axis is $M_x = \int_a^b \tilde{y} \cdot dA = \int_a^b \frac{f(x)}{2} \cdot f(x)\,dx = \int_a^b \frac{1}{2}[f(x)]^2\,dx$. Why distractors fail: Option A confuses the centroid height with a squared quantity. Option C invents an incorrect area element $f(x)^2\,dx$. Option D brings in an unrelated physics concept.

Answer: $W = \int_{-1}^{1} 800 \cdot 9.8 \cdot 5 \cdot 2\sqrt{1 - y^2}\,(1 - y)\, dy$

Determine slice geometry: For a horizontal cylinder, cross-sections perpendicular to the axis are circles. At height $y$ from center, the chord width is $2\sqrt{1 - y^2}$, and the slice has length 5 m. So the slice volume is $5 \cdot 2\sqrt{1 - y^2}\, dy$. Determine lifting distance: The top of the tank is at $y = 1$. A slice at height $y$ must be lifted $1 - y$ to exit. Assemble the integral: $W = \int_{-1}^{1} \rho g \cdot 5 \cdot 2\sqrt{1-y^2}(1-y)\,dy$, matching Option A. Why distractors fail: Option B uses $\pi(1-y^2)$ (area of a circular cross-section) instead of a chord width times length. Option C integrates only from $0$ to $1$, missing the bottom half. Option D uses $y$ instead of $(1-y)$ for the distance.

Answer: The force on the semicircle is less than the force on the rectangle, but more than half, because although narrower at depth, the wider top portions contribute significantly

Compute the rectangular force: For the rectangle: $F_{\text{rect}} = \int_0^R \rho g y \cdot 2R\, dy = \rho g R^3$. Compute the semicircular force: For the semicircle with diameter at the surface, width at depth $y$ is $2\sqrt{R^2 - y^2}$: $F_{\text{semi}} = \int_0^R \rho g y \cdot 2\sqrt{R^2 - y^2}\, dy$. Using substitution $u = R^2 - y^2$: $F_{\text{semi}} = \frac{2}{3}\rho g R^3$. Compare the results: $F_{\text{semi}}/F_{\text{rect}} = 2/3$, which is more than $1/2$ but less than $1$. The semicircle is widest near the surface (where pressure is lowest) and narrowest at depth (where pressure is highest). Why distractors fail: Option A: the ratio is $2/3$, not $1/2$. Option B: shape doesn't concentrate pressure; the semicircle is actually narrower at depth. Option C: the ratio $2/3 > 1/2$.

Answer: The work on B is exactly twice the work on A

Compute the work formula: For a spring stretched from $0$ to $d$: $W = \int_0^d kx\,dx = \frac{k d^2}{2}$. Since $d$ is the same for both, $W$ is directly proportional to $k$. Apply the comparison: $W_B / W_A = k_B / k_A = 200/100 = 2$. The work on Spring B is exactly twice that on Spring A. Why distractors fail: Option A ignores the dependence on $k$. Option C (four times) confuses proportionality to $k$ with proportionality to $k^2$. Option D ($\sqrt{2}$) has no basis in the formula.

Answer: $r(y) = y/3$

Use similar triangles: At the top ($y = 6$), the radius is 2. At the vertex ($y = 0$), the radius is 0. By similar triangles, $r/y = 2/6$, so $r(y) = y/3$. Why Option B is correct: Checking: $r(0) = 0$ (vertex) and $r(6) = 6/3 = 2$ (top), which matches the given dimensions. Why distractors fail: Option A gives $r(6) = 12$, far too large. Option C gives $r(0) = 2$ (radius at vertex should be 0). Option D gives $r(6) = 18$, vastly too large.

Answer: Yes; for a flat surface, $F = \rho g \bar{d} A$ gives the same result as integrating $\rho g y \cdot w(y)$

Verify by integration: $F = \int_2^5 9800 \cdot y \cdot 3\, dy = 29400\left[\frac{y^2}{2}\right]_2^5 = 29400 \cdot \frac{21}{2} = 308\,700$ N. Confirm the shortcut: The centroid of the rectangle is at depth $\bar{d} = (2+5)/2 = 3.5$ m. Area $A = 3 \times 3 = 9$ m². So $F = 9800 \cdot 3.5 \cdot 9 = 308\,700$ N. The shortcut $F = \rho g \bar{d} A$ is exact for any flat plate (it follows directly from the mean value theorem for integrals). Why distractors fail: Option A is wrong: the formula works for vertical (and any flat) surfaces, not just horizontal. Option B fabricates a connection to partial fractions. Option D is wrong: there is no curvature; the surface is flat, and the formula is exact.

Answer: $15$ J

Set up the work integral: $W = \int_{0.1}^{0.4} 200x\, dx$ Evaluate the integral: $W = 200 \cdot \frac{x^2}{2}\Big|_{0.1}^{0.4} = 100(0.16 - 0.01) = 100(0.15) = 15$ J. Why distractors fail: Option A ($1.5$ J) divides the correct answer by 10, likely a decimal error. Option C ($16$ J) comes from $100(0.4^2) = 16$, integrating from $0$ to $0.4$. Option D ($30$ J) doubles the correct answer, perhaps forgetting the $1/2$ from integration.

Answer: $574\,933$ N

Determine the width function: Let $y$ be depth from the surface. At $y = 0$ (top), width = 10. At $y = 4$ (bottom), width = 6. The width decreases linearly: $w(y) = 10 - \frac{10 - 6}{4}y = 10 - y$. Set up and evaluate the integral: $F = \int_0^4 9800 \cdot y \cdot (10 - y)\,dy = 9800\int_0^4(10y - y^2)\,dy = 9800\left[5y^2 - \frac{y^3}{3}\right]_0^4 = 9800\left(80 - \frac{64}{3}\right) = 9800 \cdot \frac{176}{3} = \frac{1\,724\,800}{3} \approx 574\,933$ N. Why distractors fail: Option A ($418\,133$ N) results from an arithmetic error in evaluating the integral. Option C ($470\,400$) corresponds to using a constant width of 8 m (the average): $9800 \cdot 8 \cdot \frac{16}{2} = 627\,200$—also incorrect. Option D ($627\,200$) results from using a constant width of 10 m: $9800 \cdot 10 \cdot \frac{16}{2}$, ignoring that the width varies.

Answer: $W = \frac{\delta \pi R^2 H^2}{12}$

Set up the integral: Let $y$ be measured from the vertex. The radius at height $y$ is $r(y) = Ry/H$ by similar triangles. Slice volume: $\pi(Ry/H)^2\,dy$. Distance to top: $H - y$. Evaluate: $W = \int_0^H \delta \pi \frac{R^2}{H^2} y^2 (H-y)\,dy = \frac{\delta \pi R^2}{H^2}\int_0^H (Hy^2 - y^3)\,dy$. Computing: $\int_0^H (Hy^2 - y^3)\,dy = \frac{H^4}{3} - \frac{H^4}{4} = \frac{H^4}{12}$. So $W = \frac{\delta \pi R^2}{H^2} \cdot \frac{H^4}{12} = \frac{\delta \pi R^2 H^2}{12}$. Why distractors fail: Option A ($1/6$) arises from $1/3 - 1/6$ or a similar error. Option C ($1/4$) omits the $Hy^2$ term or miscalculates. Option D ($1/3$) uses only $\int y^2\,dy$ without the $(H-y)$ factor.