Work

$F(x) = 6x^2 + 2x$ on $[1, 3]$, already in newtons and meters.

$W = \displaystyle\int_1^3 (6x^2 + 2x)\,dx$.

$\Big[2x^3 + x^2\Big]_1^3 = (54 + 9) - (2 + 1) = 60$.

$W = 60$ J.

$F(x) = \dfrac{c}{x^2}$ on $[1, 4]$.

$W = \displaystyle\int_1^4 \dfrac{c}{x^2}\,dx$.

$c\Big[-\dfrac{1}{x}\Big]_1^4 = c\left(1 - \tfrac{1}{4}\right) = \tfrac{3}{4}c$.

$\tfrac{3}{4}c = 9$, so $c = 12$.

$m = 0.25$ kg, so $F = 0.25 \times 9.8 = 2.45$ N, constant on $[0, 0.8]$ m.

$W = \displaystyle\int_0^{0.8} 2.45\,dx$.

$2.45 \times 0.8 = 1.96$.

$W = 1.96$ J.

$4$ cm $= 0.04$ m and $10$ cm $= 0.1$ m.

$12 = k(0.04)$, so $k = 300$ N/m.

From $x = 0$ to $x = 0.1$.

$W = \displaystyle\int_0^{0.1} 300x\,dx = 150(0.1)^2 = 1.5$.

$0.08$ m and $0.2$ m.

$F = 2(9.8) = 19.6$ N, so $19.6 = k(0.08)$ and $k = 245$ N/m.

From $x = 0.08$ to $x = 0.2$.

$W = \tfrac{245}{2}\left(0.2^2 - 0.08^2\right) = 122.5(0.0336) = 4.116$.

Natural length $0.3$ m; lengths $0.35$, $0.4$, $0.5$ m.

At $0.35$ m the stretch is $x = 0.05$, so $15 = k(0.05)$ and $k = 300$ N/m.

$0.4 - 0.3 = 0.1$ and $0.5 - 0.3 = 0.2$: from $x = 0.1$ to $x = 0.2$.

$W = 150\left(0.2^2 - 0.1^2\right) = 150(0.03) = 4.5$.

$y$ = height above the bottom, $0 \le y \le 4$ (the water).

$A(y) = \pi(2)^2 = 4\pi$, the same for every slice.

$1000(9.8)(4\pi)\,dy = 39\,200\pi\,dy$ N.

The spout is at $5 + 1 = 6$ m, so the slice rises $6 - y$.

$W = \displaystyle\int_0^4 39\,200\pi(6 - y)\,dy = 39\,200\pi\Big[6y - \tfrac{y^2}{2}\Big]_0^4 = 39\,200\pi(16) = 627\,200\pi$.

$y$ = height above the bottom edge, $0 \le y \le 1$.

Similar triangles: width $= 2y$ at height $y$, so $A(y) = 6(2y) = 12y$.

$1000(9.8)(12y)\,dy = 117\,600\,y\,dy$ N.

The top is at $1$ m: the slice rises $1 - y$.

$W = \displaystyle\int_0^1 117\,600\,y(1 - y)\,dy = 117\,600\left(\tfrac{1}{2} - \tfrac{1}{3}\right) = 19\,600$.

$y$ = height above the ground, $0 \le y \le 4$.

Similar triangles: $\dfrac{r}{2} = \dfrac{4 - y}{4}$, so $r = \dfrac{4 - y}{2}$ and $A(y) = \dfrac{\pi(4 - y)^2}{4}$.

$1000(9.8)\dfrac{\pi(4 - y)^2}{4}\,dy = 2450\pi(4 - y)^2\,dy$ N.

The slice rises $4 - y$.

$W = \displaystyle\int_0^4 2450\pi(4 - y)^3\,dy = 2450\pi\Big[-\tfrac{(4 - y)^4}{4}\Big]_0^4 = 2450\pi(64) = 156\,800\pi$.

$y$ = meters pulled up, $0 \le y \le 12$.

Rope left: $12 - y$ m, so $m(y) = 0.5(12 - y) + 4 = 10 - \tfrac{y}{2}$ kg.

$F(y) = 9.8\left(10 - \tfrac{y}{2}\right)$ N.

$W = 9.8\displaystyle\int_0^{12}\left(10 - \tfrac{y}{2}\right)dy = 9.8(120 - 36) = 823.2$.

$y$ = height lifted, $0 \le y \le 12$.

Time to height $y$: $t = \dfrac{y}{0.5} = 2y$ s, so $m(y) = 8 - 0.25(2y) = 8 - \tfrac{y}{2}$ kg (still $2$ kg at the top).

$F(y) = 9.8\left(8 - \tfrac{y}{2}\right)$ N.

$W = 9.8\displaystyle\int_0^{12}\left(8 - \tfrac{y}{2}\right)dy = 9.8(96 - 36) = 588$.

$y$ = meters pulled up, $0 \le y \le 4$.

Chain still hanging: $10 - y$ m, so $m(y) = 3(10 - y)$ kg.

$F(y) = 29.4(10 - y)$ N.

$W = 29.4\displaystyle\int_0^4 (10 - y)\,dy = 29.4(40 - 8) = 940.8$.