Exercises — Work — definition, dot product F·d, sign convention

Friction points opposite the motion, so the angle between force and displacement is $\theta =180°$. Since $\cos 180°=-1$, the work is negative. $W=Fd\cos 180°=-Fd<0$ What it looks like: in the left panel of figure s01 (titled "A: friction vs motion") the coral arrow (friction) and the mint arrow (motion) point in opposite directions.

For a perfect circle the pull toward the centre is perpendicular to the velocity at every instant, so $\theta =90°$ and $\cos 90°=0$. $W=Fd\cos 90°=0$ The force only steers; it never speeds up or slows down. Zero work. (Middle panel of figure s01, "B: gravity in orbit".)

Tension points up, displacement is up, so $\theta =0°$ and $\cos 0°=+1$. $W=Fd\cos 0°=+Fd>0$ Positive work — the rope pours energy into the crate. (Right panel of figure s01, "C: tension lifting".)

The displacement is horizontal, and $\theta =60°$ is the angle between the rope and that horizontal direction, so we use the horizontal share of the force $=F\cos \theta$. $W=Fd\cos \theta =50\times 8\times \cos 60°=400\times 0.5=\boxed{200\text{J}}$ Why $\cos$, not $\sin$? We want the part of the force along the road the sled travels; that is always the cosine part.

No angle is given, so use the component form. Here $\hat{ı}$ and $\hat{ȷ}$ are unit arrows pointing along $x$ and $y$; the numbers in front are $F_x,F_y$ and $d_x,d_y$. $W=F_x{d}_x+F_y{d}_y=(6)(3)+(-2)(5)=18-10=\boxed{8\text{J}}$

Gravity's magnitude is $F=mg=4\times 9.8=39.2\text{N}$, pointing down. The displacement is also down, so $\theta =0°$. We use the standard $W=Fd\cos \theta$ with $F=mg$ and $d=3\text{m}$: $W=Fd\cos 0°=(mg)d\cos 0°=39.2\times 3\times 1=\boxed{117.6\text{J}}$ (This is the origin of the familiar shortcut "$mgh$": for a straight vertical drop the displacement magnitude $d$ is the height dropped, and $\cos 0°=1$.)

(a) Applied force along motion, $\theta =0°$: $W_a=40\times 6\times \cos 0°=+240\text{J}$ (b) Friction opposes motion, $\theta =180°$: $W_f=15\times 6\times \cos 180°=-90\text{J}$ (c) Gravity has magnitude $mg=10\times 9.8=98\text{N}$, is vertical, motion horizontal, $\theta =90°$: $W_g=mg\times 6\times \cos 90°=98\times 6\times 0=0\text{J}$ (d) Normal force is vertical, motion horizontal, $\theta =90°$: $W_N=0\text{J}$. (e) Net work = sum of all works: $W_{\text{net}}=240-90+0+0=\boxed{150\text{J}}$ Why add them? Work is a scalar, so total energy delivered is just the plain sum — no vector bookkeeping. This net work is what will change the crate's kinetic energy.

Gravity points straight down with magnitude $mg=2\times 9.8=19.6\text{N}$. The displacement is $5\text{m}$ along the ramp. The angle between straight-down and the ramp's downhill direction is $\theta =90°-30°=60°$. $W_g=mgd\cos 60°=19.6\times 5\times 0.5=\boxed{49\text{J}}$ Check with height: the block drops a vertical height $h=d\sin 30°=5\times 0.5=2.5\text{m}$, and $W_g=mgh=19.6\times 2.5=49\text{J}$. Same answer — because only the vertical drop matters to gravity's work. In figure s02 the mint arrow shows the slanted path and the dashed lines show the vertical drop $h$; both routes give the same work.

The puck starts at rest, so its initial kinetic energy is zero ($v_0=0$). The theorem says the net work equals the change in kinetic energy: $W_{\text{net}}=\frac{1}{2}m{v}^2-0\Rightarrow 24=\frac{1}{2}(3)v^2=1.5v^2$ $v^2=\frac{24}{1.5}=16\Rightarrow v=\boxed{4\text{m/s}}$

Since the crate moves at steady speed, friction ($15\text{N}$) must equal the applied force — but wait, the numbers in Exercise 3.1 ($F_a=40\text{N}$, $f=15\text{N}$) are not equal, so that scenario was accelerating. For this power question we take the constant-speed case and ask only about the work done by the applied force, which is independent of whether other forces balance it. From Exercise 3.1, the applied force did $W_a=F_{a}d\cos 0°=40\times 6=240\text{J}$. Power is work per unit time: $P_{\text{avg}}=\frac{W_a}{t}=\frac{240}{3}=\boxed{80\text{W}}$ Cross-check with $P=Fv$: the crate's speed is $v=\frac{6\text{m}}{3\text{s}}=2\text{m/s}$, and $P=F_{a}v\cos 0°=40\times 2=80\text{W}$. ✓ Key distinction: power here uses the work of one specific force ($W_a$), not the net work. Net work relates to the change in kinetic energy (Exercise 4.1); the power of an individual force is $W_{\text{that force}}/t$.

Going up: displacement is up $50\text{m}$, gravity is down, $\theta =180°$: $W_{\text{up}}=mg(50)\cos 180°=-12\times 9.8\times 50=-5880\text{J}$ Coming down: displacement is down $50\text{m}$, gravity down, $\theta =0°$: $W_{\text{down}}=mg(50)\cos 0°=+5880\text{J}$ Total: $W_{\text{total}}=-5880+5880=\boxed{0\text{J}}$ Why zero? The net displacement for a round trip is zero, and gravity is a conservative force — its work depends only on start and end heights, which are the same here.

(a) Work by $\vec{F}$ using the component form (only the $x$-parts survive, since $d_y=0$): $W_F=F_x{d}_x+F_y{d}_y=(30)(4)+(-40)(0)=\boxed{120\text{J}}$ (b) The push has a downward component $F_y=-40\text{N}$, i.e. it presses down with $40\text{N}$. The floor must support both the weight and this extra downward push, so the normal force is $N=mg+|F_y|=5\times 9.8+40=49+40=89\text{N}$ Friction magnitude: $f=\mu N=0.2\times 89=17.8\text{N}$ (c) Friction opposes the motion, $\theta =180°$: $W_f=fd\cos 180°=17.8\times 4\times (-1)=\boxed{-71.2\text{J}}$ (d) Net work (gravity and normal are vertical, $\perp$ to the horizontal motion, so they do zero work): $W_{\text{net}}=120-71.2+0+0=\boxed{48.8\text{J}}$

Leg 1: $W_1=(2)(4)+(3)(0)=8\text{J}$. Leg 2: $W_2=(2)(0)+(3)(3)=9\text{J}$. Total: $W=8+9=\boxed{17\text{J}}$. Straight-line check: the overall displacement is ${\vec{d}}_1+{\vec{d}}_2=(4\hat{ı}+3\hat{ȷ})\text{m}$, so $W=(2)(4)+(3)(3)=8+9=17\text{J}.✓$ Why they match: $\vec{F}$ is constant, and the dot product distributes over vector addition: $\vec{F}\cdot ({\vec{d}}_1+{\vec{d}}_2)=\vec{F}\cdot {\vec{d}}_1+\vec{F}\cdot {\vec{d}}_2$. For a constant force, only the net displacement matters — see Dot Product (Scalar Product).

Set the ramp's downhill direction as "along motion." Gravity's along-ramp component (the part that drives it down): $W_g=mg\sin 30°\times d=4\times 9.8\times 0.5\times 10=+196\text{J}$ Normal force: perpendicular to motion, $\theta =90°$, so $W_N=0$. Friction: first the normal force is $N=mg\cos 30°=4\times 9.8\times 0.866=33.95\text{N}$, so $f=\mu N=0.25\times 33.95=8.49\text{N}$, opposing motion ($\theta =180°$): $W_f=-fd=-8.49\times 10=-84.87\text{J}$ Net work: $W_{\text{net}}=196+0-84.87=\boxed{111.13\text{J}}$ Every force is accounted for, every sign checked: gravity feeds energy in, friction drains it, the normal force does nothing.