Corrected exercises: work, energy and power (203-SN1-RE)

Vertically, the component $T\sin 30^{\circ}=50$ N relieves the floor: $N=mg-T\sin 30^{\circ}=245-50=195$ N. The normal force is not $mg$ as soon as another force has a vertical component: it is read from the $y$ equation, never by reflex.

Tension: $W_{T}=Td\cos 30^{\circ}=(100)(8)(0.866)\approx 693$ J. Friction: $f_{k}=\mu_{k}N=(0.20)(195)=39$ N, opposite to the motion, $W_{f}=-(39)(8)=-312$ J. Weight and normal force: perpendicular to the displacement, $W_{g}=W_{N}=0$.

$W_{net}=693-312\approx 381$ J. Work-energy theorem: $\frac{1}{2}mv^{2}=W_{net}$ (starting from rest), so $v=\sqrt{\frac{2(381)}{25}}\approx 5.5$ m/s.

Not necessarily, and here it does not: pulling horizontally raises the work of the tension to $Td=800$ J, but removes the vertical relief: $N$ goes back up to 245 N, $f_{k}$ to 49 N, and $W_{f}$ to $-392$ J. Balance: $W_{net}=800-392=408$ J, barely better than 381 J. The two effects work in opposite directions; with a larger $\mu_{k}$, pulling upward at an angle can even become the better choice. That is the whole point of the complete balance: no single force decides anything.

Because $F$ is not constant: it changes at every position. The formula $W=Fd\cos\theta$ assumes a force of constant magnitude and direction. The displacement must be cut into slices $dx$ thin enough for $F$ to be nearly constant on each, then summed: $W=\int_{x_{i}}^{x_{f}}F(x)\,dx$. Graphically, this integral is exactly the AREA under the curve $F(x)$, counted positive above the axis and negative below it. It is the same idea as the area under a velocity-time graph giving the displacement.

First interval, a triangle: $W_{1}=\frac{1}{2}(3.00)(12.0)=18.0$ J. Second, a rectangle: $W_{2}=(7.00-3.00)(12.0)=48.0$ J. Third, a triangle: $W_{3}=\frac{1}{2}(3.00)(12.0)=18.0$ J. Total: $W=18.0+48.0+18.0=84.0$ J. No antiderivative was needed: cutting the area into geometric shapes is enough as soon as the curve is made of straight segments.

There is no friction and the motion is horizontal, so the net work reduces to that of $F$. Work-energy theorem: $\frac{1}{2}mv^{2}=W$, that is $v=\sqrt{\frac{2W}{m}}$. At $x=3.00$ m: $v=\sqrt{\frac{2(18.0)}{4.00}}=3.00$ m/s. At $x=7.00$ m ($W=66.0$ J): $v=\sqrt{33.0}\approx 5.74$ m/s. At $x=10.0$ m ($W=84.0$ J): $v=\sqrt{42.0}\approx 6.48$ m/s. The speed is greatest at the end, at $x=10.0$ m: the force stays positive everywhere, so it never stops adding kinetic energy, even when it weakens. A decreasing force does not brake, it just accelerates less and less quickly.

The force that must be exerted to compress the spring is $F=kx$ (opposite to the restoring force). So $W=\int_{0}^{x}kx'\,dx'=\frac{1}{2}kx^{2}$. For $k=800$ N/m and $x=0.150$ m: $W=\frac{1}{2}(800)(0.150)^{2}=9.00$ J. From $0.150$ m to $0.300$ m: $W=\frac{1}{2}(800)\left[(0.300)^{2}-(0.150)^{2}\right]=400(0.0675)=27.0$ J, that is THREE times as much for the same extra compression. The reason is geometric: on the $F(x)$ graph, the second slice is a trapezoid much taller than the first triangle. That is why the last centimetres of a spring, a bow or a shock absorber always cost the most.

Friction opposes the motion over the whole path: $W_{f}=-(5.00)(10.0)=-50.0$ J. Net work: $84.0-50.0=34.0$ J, a drop of $59.5\,\%$. Final speed: $v=\sqrt{\frac{2(34.0)}{4.00}}=\sqrt{17.0}\approx 4.12$ m/s, against $6.48$ m/s without friction, a drop of only $36.4\,\%$. The two proportions differ because the speed varies as the SQUARE ROOT of the work: $v\propto\sqrt{W}$. Dividing the energy by four divides the speed by only two. It is the same effect, taken in reverse, as the stopping distance that quadruples when the speed doubles.

Braking removes all the kinetic energy: $Fd=\frac{1}{2}mv^{2}$, so $d=\frac{(0.5)(1200)(25)^{2}}{7500}=\frac{375\,000}{7500}=50$ m.

$d=\frac{mv^{2}}{2F}$ is proportional to $v^{2}$: doubling the speed quadruples the distance, $d=200$ m. This is the heart of road safety campaigns: at 180 km/h you need four times as much road as at 90 km/h, not twice as much.

The acceleration is $a=\frac{F}{m}=6.25$ m/s², so $t=\frac{v}{a}=\frac{25}{6.25}=4$ s. The time only doubles (8 s at 180 km/h): time depends on $v$ (the speed must be cancelled, at the constant rate $a$), while distance depends on $v^{2}$ (the energy must be dissipated, and it grows as the square). Two quantities, two laws.

The theorem links the initial state (25 m/s) directly to the final state (stopped) without going through time, which is unknown here: one equation, one unknown. Kinematic check: $0=(25)^{2}-2(6.25)\Delta x$ does give $\Delta x=\frac{625}{12.5}=50$ m. The two methods agree; energy simply avoids the detour.

The tension is at every instant perpendicular to the velocity (the string is radial, the motion tangential): $W=Fd\cos 90^{\circ}=0$. The only force that does work is the weight, a conservative force: the mechanical energy is conserved.

Height: $h=L(1-\cos\theta_{0})=2.5(1-0.5)=1.25$ m. Conservation: $\frac{1}{2}mv^{2}=mgh$, so $v=\sqrt{2(9.8)(1.25)}=\sqrt{24.5}\approx 4.95$ m/s.

At the lowest point, the bob moves along a circle: the net force must point toward the centre (upward) with magnitude $\frac{mv^{2}}{L}$. So $T-mg=\frac{mv^{2}}{L}$, that is $T=0.8(9.8)+\frac{0.8(24.5)}{2.5}=7.84+7.84=15.68$ N. The tension exceeds the weight because it must not only support the bob but also bend its path: the excess is exactly the centripetal force.

Conservation from the angle $\theta_{0}$: $v^{2}=2gL(1-\cos\theta_{0})$. At the bottom: $T=mg+\frac{mv^{2}}{L}=mg+2mg(1-\cos\theta_{0})=mg(3-2\cos\theta_{0})$. The length of the string has dropped out of the result. At $\theta_{0}=60^{\circ}$: $T=mg(3-1)=2mg$, which does give 15.68 N again for our bob. $T>2mg$ requires $3-2\cos\theta_{0}>2$, that is $\cos\theta_{0}<\frac{1}{2}$: beyond 60°. Released from the horizontal ($\theta_{0}=90^{\circ}$), the string supports $3mg$: this is the strength criterion to remember when sizing a cable.

$E_{e}=\frac{1}{2}kx^{2}=\frac{1}{2}(200)(0.10)^{2}=1.00$ J. On a smooth surface, everything goes into kinetic energy: $v=\sqrt{\frac{2(1.00)}{0.5}}=2.0$ m/s.

Friction: $f_{k}=\mu_{k}mg=(0.30)(0.5)(9.8)=1.47$ N. Energy dissipated: $f_{k}d=(1.47)(0.50)=0.735$ J. That leaves $1.00-0.735=0.265$ J, so $v=\sqrt{\frac{2(0.265)}{0.5}}\approx 1.03$ m/s.

The block stops when all the energy has been dissipated: $d_{max}=\frac{E}{f_{k}}=\frac{1.00}{1.47}\approx 0.68$ m. The 50 cm patch is crossed, but only just: 18 cm more of rough surface would be enough to stop the block. Three quarters of the initial energy ($0.735$ out of $1.00$ J) stays in the rough patch.

Balance: elastic potential energy (1.00 J) $\rightarrow$ kinetic energy (1.00 J) $\rightarrow$ kinetic energy (0.265 J) $+$ energy dissipated by friction (0.735 J), then nothing more changes on the final smooth surface. The dissipated energy becomes thermal energy: the surfaces in contact warm up slightly. It is not destroyed, but it can no longer be recovered as motion: that is the mark of a non-conservative force.

At equilibrium, the restoring force exactly balances the weight: $kx_{eq}=mg$, so $x_{eq}=\frac{mg}{k}=\frac{2.00(9.80)}{250}=0.0784$ m, or $7.84$ cm.

Here we must conserve energy, not balance forces. The block starts from rest and stops at the lowest point, so $\Delta K=0$: all the gravitational potential energy lost has been stored in the spring. $mgx_{max}=\frac{1}{2}kx_{max}^{2}$, whence $x_{max}=\frac{2mg}{k}=0.157$ m, or $15.7$ cm. That is EXACTLY double the equilibrium extension, and the result is general: $x_{max}=2x_{eq}$, whatever the mass and the stiffness. A load released suddenly stresses a structure twice as much as a load lowered gently, which is a basic principle of engineering design.

The confusion comes from equating zero force with zero velocity, whereas force governs the CHANGE of velocity, not the velocity itself. At the lowest point, the spring pulls harder than the weight: the net force points upward, it is even at its maximum, and this is precisely what sends the block back up. Its speed is zero there for an instant, but it is changing very fast. At the equilibrium position, on the contrary, the net force vanishes: the speed stops changing, so it passes through its maximum. Maximum force and zero speed go together; zero force and maximum speed go together too.

The speed is greatest at the equilibrium position, $x_{eq}=0.0784$ m. Energy balance from the release: $\frac{1}{2}mv^{2}=mgx_{eq}-\frac{1}{2}kx_{eq}^{2}=2.00(9.80)(0.0784)-\frac{1}{2}(250)(0.0784)^{2}=1.537-0.768=0.768$ J. So $v=\sqrt{\frac{2(0.768)}{2.00}}\approx 0.877$ m/s. Second method: the motion is a simple harmonic oscillation of amplitude $x_{eq}$ about the equilibrium position, and $v_{max}=x_{eq}\sqrt{\frac{k}{m}}=0.0784\sqrt{125}\approx 0.877$ m/s. The two methods agree. Note in passing that the energy stored at equilibrium is exactly half the gravitational energy lost, the other half being the kinetic energy.

Take the axis pointing down, the origin at the natural length, and set $x=u+x_{eq}$. The total potential energy is $U=-mgx+\frac{1}{2}kx^{2}$ (the minus sign because going down loses gravitational energy). Substituting: $U=-mg\left(u+x_{eq}\right)+\frac{1}{2}k\left(u+x_{eq}\right)^{2}=-mgu-mgx_{eq}+\frac{1}{2}ku^{2}+kx_{eq}u+\frac{1}{2}kx_{eq}^{2}$. Now $kx_{eq}=mg$, so the linear terms $-mgu$ and $+kx_{eq}u$ cancel, leaving $U=\frac{1}{2}ku^{2}+\left(\frac{1}{2}kx_{eq}^{2}-mgx_{eq}\right)=\frac{1}{2}ku^{2}-\frac{1}{2}kx_{eq}^{2}$. The constant has no influence since only changes in $U$ matter. Consequence: a VERTICAL mass-spring system behaves exactly like a HORIZONTAL mass-spring system, provided positions are measured from the equilibrium position. Gravity merely shifts the equilibrium point; it changes neither the period nor the form of the motion. This is the result that will let you treat both cases with a single formula in the chapter on simple harmonic motion.

Constant speed: zero acceleration, so $\Sigma F=0$ along the slope. The driving force must balance the component of the weight AND the resistances: $F_{tr}=mg\sin 5^{\circ}+600=(1500)(9.8)(0.0872)+600\approx 1281+600=1881$ N.

$P=F_{tr}\,v=(1881)(20)\approx 37\,600$ W $\approx 37.6$ kW. (For reference, about 50 horsepower: climbing a hill at a good pace uses a large share of the power of an ordinary car.)

Duration: $t=\frac{2000}{20}=100$ s, so $E=Pt\approx 3.76$ MJ. Gain in altitude: $h=2000\sin 5^{\circ}\approx 174.3$ m, so $\Delta E_{p}=mgh\approx (1500)(9.8)(174.3)\approx 2.56$ MJ. Dissipation: $(600)(2000)=1.20$ MJ. Check: $2.56+1.20=3.76$ MJ, the books balance. About two thirds of the fuel energy used for the climb ends up stored as altitude.

On the level, only the 600 N resistance remains to be overcome: $P=(600)(20)=12$ kW, barely a third of the climbing power. The difference, $25.6$ kW, is exactly the rate at which potential energy is gained, $mgv\sin 5^{\circ}$: climbing is essentially paying gravity in real time.

On the way down, the component of the weight ($\approx 1281$ N) pushes the car and exceeds the resistances (600 N): without brakes, the car would accelerate. To hold 20 m/s, the brakes must dissipate the difference: $F_{fr}=1281-600=681$ N, that is $P_{fr}=(681)(20)\approx 13.6$ kW. This is why long descents overheat brakes and why engine braking is recommended: 13.6 kW dissipated continuously is the equivalent of about ten kettles.

At the top, two forces act on the cart: its weight $mg$ and the normal force $N$ of the rail, both directed DOWNWARD, that is, toward the centre of the loop. Newton's second law along the radial direction gives $mg+N=\frac{mv^{2}}{R}$. A rail can only push, never pull, so $N\geq 0$. The limiting case is $N=0$, where gravity alone supplies all the centripetal acceleration: $v_{min}=\sqrt{gR}=\sqrt{9.80(2.50)}\approx 4.95$ m/s. Below that, the cart leaves the rail and becomes a projectile.

Without friction, mechanical energy is conserved between the release point (height $h$, at rest) and the top of the loop (height $2R$, speed $v_{min}$): $mgh=mg(2R)+\frac{1}{2}mv_{min}^{2}$. Substituting $v_{min}^{2}=gR$: $gh=2gR+\frac{gR}{2}$, so $h_{min}=\frac{5R}{2}=2.5R$. Numerically $h_{min}=2.5(2.50)=6.25$ m, that is $1.25$ m above the top of the loop. The mass cancels everywhere: a loaded cart and an empty cart complete the loop from the same height, which is fortunate for the operator of an amusement park.

Conservation of energy between $8.00$ m and the top, which is at $2R=5.00$ m: $\frac{1}{2}v^{2}=g(8.00-5.00)$, so $v^{2}=2(9.80)(3.00)=58.8$ and $v\approx 7.67$ m/s, well above the $4.95$ m/s required. Force of the rail: $N=\frac{mv^{2}}{R}-mg=\frac{0.500(58.8)}{2.50}-0.500(9.80)=11.76-4.90=6.86$ N, directed downward, that is, toward the centre.

At the lowest point, the weight points down and the normal force up, that is, toward the centre: $N-mg=\frac{mv^{2}}{R}$. Here $v^{2}=2g(8.00)=156.8$ m$^{2}$/s$^{2}$, so $N=\frac{0.500(156.8)}{2.50}+4.90=31.36+4.90=36.3$ N. That is $7.4$ times the weight of the cart. A passenger there would feel an apparent weight $7.4$ times their own: beyond $5$ to $6\,g$ peripheral vision is lost and fainting threatens, which is precisely the limit designers refuse to cross.

Only $75.0\,\%$ of the initial energy is left at the entrance of the loop, and the loop itself is assumed smooth. So we need $0.750\,mgh\geq mg(2.5R)$, that is $h\geq\frac{2.5(2.50)}{0.750}=8.33$ m. Friction costs more than two extra metres of height. As for the shape: a circular loop demands an enormous speed at the bottom to hold on at the top, hence a centripetal acceleration $\frac{v^{2}}{R}$ that is excessive at the entrance, exactly the problem of part d). Designers use a clothoid, a curve whose radius of curvature is large at the bottom, where the speed is high, and small at the top, where it is low. The acceleration $\frac{v^{2}}{R}$ then stays roughly constant and bearable along the whole ride, and the release height needed goes down.

Conservation on the icy part: $mgh=\frac{1}{2}mv^{2}$, so $v=\sqrt{2(9.8)(8)}\approx 12.5$ m/s. The mass cancels out of the equation. The shape of the profile does not matter because the weight is conservative: its work depends only on the height difference $h$, not on the path; and the normal force, always perpendicular to the motion, does no work on any profile. Only the start and the finish count.

On the flat, friction $\mu_{k}mg$ dissipates the energy $mgh$ gained: $\mu_{k}mg\,d=mgh$, so $d=\frac{h}{\mu_{k}}=\frac{8}{0.25}=32$ m. The mass AND $g$ have disappeared: a heavy or a light sled, on Earth or on the Moon, stops at the same distance. All that remains is the geometry ($h$) and the quality of the surface ($\mu_{k}$).

The 12 m of snow take away the equivalent of $\mu_{k}\times 12=3$ m of height. Balance in equivalent heights: $h'=8-(0.25)(12)=5$ m. The sled climbs back to exactly 5 m (the second hill is icy: no loss on the way up).

The sled slides back down 5 m, recrosses the 12 m of rough snow (another loss of 3 m equivalent) and reaches the first hill with $5-3=2$ m equivalent: it climbs to 2 m, then comes back down and sets off over the flat with only 2 m equivalent. It can no longer cross: it travels $\frac{2}{0.25}=8$ m and stops for good INSIDE the rough zone, two thirds of the way across (4 m from the second hill). Overall check: the sled has travelled $12+12+8=32$ m of rough snow, dissipating $(0.25)(32)=8$ m equivalent, exactly the initial height. All the energy $mgh$ has ended up as heat in the snow.

The force derives from the potential: $F(x)=-\frac{dU}{dx}=-\left(4x^{3}-16x\right)=-4x\left(x^{2}-4\right)$. It vanishes at $x=0$ and $x=\pm 2.00$ m, the three equilibrium positions. At $x=0$, $U(0)=0$ is a local MAXIMUM: UNSTABLE equilibrium. The force confirms it: for $x$ slightly positive, $F=-4x(x^{2}-4)>0$, so the force pushes even farther from the origin; the displacement grows. At $x=\pm 2.00$ m, $U=16-32=-16.0$ J is a MINIMUM: STABLE equilibrium. For $x$ a little above $2$, $F<0$ and brings the particle back toward $2$; for $x$ a little below, $F>0$ and brings it back as well. General rule: the force always points toward decreasing $U$, so a dip restores and a hump repels.