Corrected exercises: work (MATH 141)

The force is constant and along the motion, so $W = Fd = 12 \times 5 = 60$ J. As an integral, $W = \int_0^5 12\,dx = 12x \Big|_0^5 = 60$ J: the area of the rectangle of width $5$ (the path) and height $12$ (the force) under the graph of $F$. This is the picture to keep for the whole chapter: work is an AREA under a force-position graph, and the product $Fd$ is only the special case where that area is a rectangle.

Cut $[1, 3]$ into pieces of length $\Delta x$. On one piece the force is almost $F(x_i^*)$, so the work on it is almost $F(x_i^*)\Delta x$, and the Riemann sum $\sum F(x_i^*)\Delta x$ tends to $\int_1^3 F(x)\,dx$. Here $W = \int_1^3 (3x^2 + 2x)\,dx = \left[x^3 + x^2\right]_1^3 = (27 + 9) - (1 + 1) = 34$ J. The two products are $F(1) \times 2 = 5 \times 2 = 10$ and $F(3) \times 2 = 33 \times 2 = 66$. Since $F$ increases on $[1, 3]$, it stays between $5$ and $33$, and the comparison property of the integral gives $10 \le W \le 66$. The check catches the classic slip, multiplying the FINAL force by the distance ($66$ J), which pretends the force was $33$ N all the way.

$W = \int_1^3 6x^{-2}\,dx = \left[-\frac{6}{x}\right]_1^3 = -2 - (-6) = 4$ J. The force now decreases from $6$ N to $\frac{2}{3}$ N, so this time the bracket is $\frac{2}{3} \times 2 \le W \le 6 \times 2$, that is $\frac{4}{3} \le 4 \le 12$: consistent. The minus sign of the antiderivative $-\frac{6}{x}$ is where copies go wrong; a negative work for a force that pushes forward the whole way is the signal.

The work is the SIGNED area between the graph and the $x$-axis, read with geometry. On $[0, 2]$: triangle, $\frac{1}{2} \times 2 \times 4 = 4$. On $[2, 5]$: rectangle, $3 \times 4 = 12$. So $W_{0 \to 5} = 16$ J. On $[5, 8]$ the line goes from $4$ down to $-2$, with slope $-2$, so it crosses zero at $x = 7$: triangle above the axis on $[5, 7]$, area $\frac{1}{2} \times 2 \times 4 = 4$, and triangle BELOW the axis on $[7, 8]$, area $\frac{1}{2} \times 1 \times 2 = 1$, counted negative because the force now points backwards. $W_{0 \to 8} = 16 + 4 - 1 = 19$ J. The accumulated work $W(b) = \int_0^b F(x)\,dx$ has derivative $W'(b) = F(b)$ by the Fundamental Theorem: it increases while $F > 0$ and decreases once $F < 0$, so it is largest at $x = 7$, where $W = 20$ J. Adding the last triangle as $+1$ ($21$ J) is the trap: work done against the motion is negative.

With $u = \frac{\pi x}{4}$, $du = \frac{\pi}{4}\,dx$: $\int 10\sin\left(\frac{\pi x}{4}\right)dx = -\frac{40}{\pi}\cos\left(\frac{\pi x}{4}\right) + C$. From $0$ to $4$: $-\frac{40}{\pi}(\cos\pi - \cos 0) = -\frac{40}{\pi}(-2) = \frac{80}{\pi}$ J, about $25$ J with $\pi \approx 3.14$. From $0$ to $8$: $-\frac{40}{\pi}(\cos 2\pi - \cos 0) = 0$. The force pushes forward on $[0, 4]$ and backward on $[4, 8]$, symmetrically, so the net work is zero even though the particle moved $8$ m. Forgetting the factor $\frac{4}{\pi}$ of the substitution gives $20$ J instead of $\frac{80}{\pi}$: the result is off by the factor $\frac{4}{\pi}$, which is not $1$.

$40 = k(0.2)$ gives $k = 200$ N/m. Then $W = \int_0^{0.2} 200x\,dx = 100x^2 \Big|_0^{0.2} = 100 \times 0.04 = 4$ J. The classmate's $8$ J multiplies the force at the END of the stretch by the whole distance, as if the spring had resisted with $40$ N from the start. It did not: the force grows from $0$ to $40$ N, so the area under $F = 200x$ is a TRIANGLE, $\frac{1}{2} \times 0.2 \times 40 = 4$, exactly half the rectangle $40 \times 0.2$. On an exam this answer loses the whole question, because it skips the integral that the chapter is about.

$W = \int_{0.2}^{0.3} 200x\,dx = 100(0.3^2 - 0.2^2) = 100(0.09 - 0.04) = 5$ J, while $\int_0^{0.1} 200x\,dx = 100 \times 0.01 = 1$ J. Same displacement of $0.1$ m, five times the work: the second stretch starts where the spring already pulls back with $40$ N. The bounds of a spring integral are STRETCHES, and the work between two of them is $\frac{k}{2}(x_2^2 - x_1^2)$, not $\frac{k}{2}(x_2 - x_1)^2$, which would give the wrong $1$ J here.

The work from the natural length to a stretch $x$ is $\int_0^x 200t\,dt = 100x^2$. Setting $100x^2 = 16$ gives $x^2 = 0.16$, so $x = 0.4$ m, the positive root since $x$ is a stretch. Four times the work for twice the stretch: the work grows like the SQUARE of the stretch.

Let $L$ be the natural length in metres. The lengths $0.30$, $0.40$, $0.50$ correspond to the stretches $0.30 - L$, $0.40 - L$, $0.50 - L$. Then $1.5 = \frac{k}{2}\left[(0.4 - L)^2 - (0.3 - L)^2\right] = \frac{k}{2}(0.1)(0.7 - 2L)$ and $2.5 = \frac{k}{2}(0.1)(0.9 - 2L)$, by the difference of squares. Dividing the second by the first: $\frac{0.9 - 2L}{0.7 - 2L} = \frac{5}{3}$, so $2.7 - 6L = 3.5 - 10L$, $4L = 0.8$ and $L = 0.2$ m $= 20$ cm. Back in the first equation: $0.05k \times 0.3 = 1.5$, so $k = 100$ N/m. Check the second: $0.05 \times 100 \times 0.5 = 2.5$ J. The trap is to integrate from $0.30$ to $0.40$, which treats the LENGTH as the stretch and makes the spring natural at length zero.

A length of $15$ cm is a compression of $0.05$ m below the natural $20$ cm. Hooke's law holds in compression with the same $k$, so $W = \int_0^{0.05} 100x\,dx = 50 \times 0.0025 = 0.125 = \frac{1}{8}$ J. The work YOU do is positive whether you stretch or compress: you push in the direction the end moves.

The slice between depths $x$ and $x + \Delta x$ has length $\Delta x$, so it weighs $6\,\Delta x$ newtons, and it must rise about $x$ metres to reach the roof. Its work is about $6x\,\Delta x$. Adding the slices gives a Riemann sum $\sum 6x_i^*\,\Delta x$, whose limit is $W = \int_0^{20} 6x\,dx = 3x^2 \Big|_0^{20} = 1200$ J. The famous wrong answer is $120 \times 20 = 2400$ J: total weight times total length, which lifts EVERY piece $20$ m. Check with the average value: the cable is uniform, every metre weighs the same, and the distances $x$ run evenly from $0$ to $20$, with average $\frac{1}{20}\int_0^{20} x\,dx = 10$ m. So $W = 120 \times 10 = 1200$ J.

Two groups of slices. The top $10$ m end on the roof: a slice at depth $x \in [0, 10]$ rises $x$, total $\int_0^{10} 6x\,dx = 3x^2 \Big|_0^{10} = 300$ J. The bottom $10$ m, weighing $60$ N, are NOT wound up but they rise too: every piece of it moves up exactly $10$ m, total $60 \times 10 = 600$ J. So $W = 900$ J. Answering $300$ J forgets that the lower half is dragged up with the upper half, and it is the most frequent error on this question.

The crate is a single weight of $100$ N that rises the full $20$ m: constant force, no integral, $100 \times 20 = 2000$ J. The cable still costs $1200$ J, so $W = 3200$ J. Integrating the crate, or forgetting it, both cost the part.

After $y$ metres have been wound, $20 - y$ metres still hang, weighing $6(20 - y)$ newtons, plus the crate. The winch pulls with $F(y) = 6(20 - y) + 100 = 220 - 6y$ newtons, a force that VARIES along the path, so $W = \int_0^{20} (220 - 6y)\,dy = \left[220y - 3y^2\right]_0^{20} = 4400 - 1200 = 3200$ J, as in c). Two readings of the same work: slice the object and add weight times distance, or slice the path and add force times displacement. Choose the one where the two factors are easy to write.

The piece of chain that ends at height $y$ was lying on the ground, at height $0$, so it rises $y$: a slice of length $\Delta y$ weighs $8\,\Delta y$ and costs $8y\,\Delta y$. $W = \int_0^{10} 8y\,dy = 4y^2 \Big|_0^{10} = 400$ J. When the end is lifted to $4$ m only, only $4$ m of chain leave the ground, and the other $6$ m do not move: $W = \int_0^4 8y\,dy = 4 \times 16 = 64$ J. The bounds follow the pieces that MOVE, not the length of the chain.

Every layer is a disc of radius $2$: $A(y) = 4\pi$, volume $4\pi\,\Delta y$, weight $9800 \times 4\pi\,\Delta y = 39\,200\pi\,\Delta y$ newtons. A layer at height $y$ must reach the top, at height $5$: it travels $5 - y$. So $W = \int_0^5 39\,200\pi(5 - y)\,dy = 39\,200\pi\left[5y - \frac{y^2}{2}\right]_0^5 = 39\,200\pi \times \frac{25}{2} = 490\,000\pi$ J, about $1.5 \times 10^6$ J. The distance is $5 - y$, not $y$: $y$ is where the layer IS, $5 - y$ is how far it GOES. For this tank, $\int_0^5 y\,dy$ also equals $\frac{25}{2}$ by coincidence, which is why the error survives on a cylinder and costs the whole question on any other shape (Exercise 5).

Only the destination changes: the outlet is at height $6$, so $D(y) = 6 - y$. $W = 39\,200\pi\int_0^5 (6 - y)\,dy = 39\,200\pi\left(30 - \frac{25}{2}\right) = 39\,200\pi \times \frac{35}{2} = 686\,000\pi$ J. Check: the extra $196\,000\pi$ J is the total weight of the water, $39\,200\pi \times 5 = 196\,000\pi$ N, lifted one more metre. Every layer travels $1$ m further, so the work grows by weight times $1$ m.

The bounds are where the water IS: $y$ from $0$ to $3$. The destination is still the top of the tank, so the distance is still $5 - y$. $W = 39\,200\pi\int_0^3 (5 - y)\,dy = 39\,200\pi\left(15 - \frac{9}{2}\right) = 39\,200\pi \times \frac{21}{2} = 411\,600\pi$ J. Two errors are waiting here: $3 - y$ as the distance (the layer must reach the TOP of the tank, not the surface of the water), and bounds from $2$ to $5$ (the empty part of the tank holds nothing to pump).

The water that is removed is the water between heights $2$ and $5$; the bottom $2$ m never move. $W = 39\,200\pi\int_2^5 (5 - y)\,dy = 39\,200\pi\left[5y - \frac{y^2}{2}\right]_2^5 = 39\,200\pi\left(\frac{25}{2} - 8\right) = 39\,200\pi \times \frac{9}{2} = 176\,400\pi$ J. Writing $\int_0^3$ here computes c) again: it pumps the bottom $3$ m of water, the heaviest part of the job, which is exactly the water that stays.

In a cylinder every layer of thickness $\Delta y$ has the SAME weight, so the work is the total weight times the average distance, the average value of $D$ on the interval where the water is. For a), total weight $39\,200\pi \times 5 = 196\,000\pi$ N and average of $5 - y$ on $[0, 5]$ equal to $\frac{1}{5}\int_0^5 (5 - y)\,dy = 2.5$ m: $196\,000\pi \times 2.5 = 490\,000\pi$ J. For c), weight $39\,200\pi \times 3 = 117\,600\pi$ N and average of $5 - y$ on $[0, 3]$ equal to $\frac{1}{3} \times \frac{21}{2} = 3.5$ m: $117\,600\pi \times 3.5 = 411\,600\pi$ J. The shortcut rests on the equal weights; the moment the cross-section changes with $y$, it breaks, as Exercise 5 shows.

Cut the cone by a vertical plane through its axis: the vertex, the axis and the rim make a right triangle with legs $6$ (height) and $3$ (radius). The layer at height $y$ above the vertex cuts a smaller, similar triangle with legs $y$ and $r$, so $\frac{r}{y} = \frac{3}{6}$ and $r = \frac{y}{2}$. The layer's volume is $\pi\left(\frac{y}{2}\right)^2\Delta y = \frac{\pi}{4}y^2\,\Delta y$ and its weight $9800 \times \frac{\pi}{4}y^2\,\Delta y = 2450\pi y^2\,\Delta y$ newtons. Writing $r = 2y$ (the ratio upside down) or $r = 3$ (the rim radius for every layer) is the classic loss; test the formula at $y = 6$, where it must give $3$.

The rim is at height $6$, so the layer travels $6 - y$. $W = \int_0^6 2450\pi y^2(6 - y)\,dy = 2450\pi\left[2y^3 - \frac{y^4}{4}\right]_0^6 = 2450\pi(432 - 324) = 2450\pi \times 108 = 264\,600\pi$ J. With the reflex distance $y$ instead, $2450\pi\int_0^6 y^3\,dy = 2450\pi \times 324$, three times too much: here the coincidence of the cylinder does not happen.

The bounds are where the water is, $0$ to $4$; the destination is still the rim. $W = 2450\pi\left[2y^3 - \frac{y^4}{4}\right]_0^4 = 2450\pi(128 - 64) = 2450\pi \times 64 = 156\,800\pi$ J. The water is $\frac{2}{3}$ of the height but only $\left(\frac{2}{3}\right)^3 = \frac{8}{27}$ of the volume, and still more than half of the work of b), because it sits where the distance to the rim is longest.

Vertex up, the radius SHRINKS with the height: at height $y$ the similar triangle has legs $6 - y$ and $r$, so $r = \frac{6 - y}{2}$, and the vertex is at height $6$, so the distance is again $6 - y$. $W = \int_0^6 9800\pi\frac{(6 - y)^2}{4}(6 - y)\,dy = 2450\pi\int_0^6 (6 - y)^3\,dy = 2450\pi\left[-\frac{(6 - y)^4}{4}\right]_0^6 = 2450\pi \times 324 = 793\,800\pi$ J, exactly three times b). The volume is the same, $\frac{1}{3}\pi(3)^2(6) = 18\pi$ cubic metres, but vertex up the wide layers are at the bottom, far from the outlet, and vertex down they are at the top, close to it. The work depends on WHERE the weight is, not only on how much there is.

Total weight: $9800 \times 18\pi = 176\,400\pi$ N; average of $6 - y$ on $[0, 6]$: $3$ m; product $529\,200\pi$ J, twice the true $264\,600\pi$ J. The average value gives every height the same importance, which is right only when every layer weighs the same (the cylinder). In the vertex-down cone, the heavy layers are high up, where the distance is short, so the true work weights the short distances more: dividing, the effective distance is $\frac{264\,600\pi}{176\,400\pi} = 1.5$ m, not $3$ m. The integral $\int \rho g\,A(y)D(y)\,dy$ does that weighting automatically; the shortcut does not.

The end view is a triangle with its vertex at the bottom: at height $2$ the width is $4$, at height $0$ it is $0$, and the width grows linearly with the height. By similar triangles, $\frac{w}{y} = \frac{4}{2}$, so $w = 2y$. The layer at height $y$ is a thin rectangular slab, $10$ ft long, $2y$ ft wide and $\Delta y$ ft thick: volume $20y\,\Delta y$ ft$^3$, weight $62.5 \times 20y\,\Delta y = 1250y\,\Delta y$ lb. Test: at $y = 2$ the width is $4$, as drawn.

The top edge is at height $2$, so the layer travels $2 - y$. $W = \int_0^2 1250y(2 - y)\,dy = 1250\left[y^2 - \frac{y^3}{3}\right]_0^2 = 1250\left(4 - \frac{8}{3}\right) = 1250 \times \frac{4}{3} = \frac{5000}{3}$ ft-lb, about $1667$ ft-lb.

Bounds $0$ to $1$, distance still $2 - y$: $W = 1250\left[y^2 - \frac{y^3}{3}\right]_0^1 = 1250 \times \frac{2}{3} = \frac{2500}{3}$ ft-lb. The full trough holds $10 \times \frac{1}{2} \times 4 \times 2 = 40$ ft$^3$; at a depth of $1$ ft the triangle of water is $2$ ft across and $1$ ft deep, so it holds $10 \times \frac{1}{2} \times 2 \times 1 = 10$ ft$^3$: a QUARTER of the water. Yet it costs HALF the work, because that water is the bottom of the trough, the part that travels farthest. Half the depth, a quarter of the water, half the work: three different fractions, and none of them is the one you would guess.

The pipe is at height $3$: $W = \int_0^2 1250y(3 - y)\,dy = 1250\left[\frac{3y^2}{2} - \frac{y^3}{3}\right]_0^2 = 1250\left(6 - \frac{8}{3}\right) = 1250 \times \frac{10}{3} = \frac{12\,500}{3}$ ft-lb. One-line check: every layer travels $1$ ft more than in b), so the work grows by the total weight, $62.5 \times 40 = 2500$ lb, times $1$ ft: $\frac{5000}{3} + 2500 = \frac{12\,500}{3}$.

The pound is a force, and $62.5$ lb/ft$^3$ is the WEIGHT of a cubic foot of water, the imperial analogue of $\rho g$, not of $\rho$. Multiplying by $32$ ft/s$^2$ applies the gravity a second time: the units become lb$\cdot$ft/s$^2$ per cubic foot, which is not a weight density, and every work comes out $32$ times too large, $\frac{160\,000}{3}$ ft-lb for b). In SI the density $1000$ kg/m$^3$ is a MASS density and needs $g$; in imperial the data is already a weight. Read the unit of the data before deciding whether $g$ enters.

The layer at depth $y$ is a disc. Its edge is on the sphere, at distance $3$ from the centre, which sits in the middle of the rim; the depth $y$ and the radius $r$ of the disc are the two legs of a right triangle with hypotenuse $3$. So $r^2 = 9 - y^2$ and $A(y) = \pi(9 - y^2)$. With the axis pointing down from the rim, the distance to the rim IS $y$. $W = \int_0^3 9800\pi(9 - y^2)y\,dy = 9800\pi\left[\frac{9y^2}{2} - \frac{y^4}{4}\right]_0^3 = 9800\pi\left(\frac{81}{2} - \frac{81}{4}\right) = 9800\pi \times \frac{81}{4} = 198\,450\pi$ J.

Upward from the bottom, the centre of the sphere is at height $3$, so a layer at height $y$ is $3 - y$ below the centre: $r^2 = 9 - (3 - y)^2 = 6y - y^2$, and the distance to the rim is $3 - y$. $\int_0^3 (6y - y^2)(3 - y)\,dy = \int_0^3 (18y - 9y^2 + y^3)\,dy = \left[9y^2 - 3y^3 + \frac{y^4}{4}\right]_0^3 = 81 - 81 + \frac{81}{4} = \frac{81}{4}$, the same $W = 198\,450\pi$ J. The physics cannot depend on the axis, and it does not; the algebra does. The first choice puts the origin where the geometry is centred, and it is the one to take on an exam.