Corrected exercises: work, energy and power (PHYS 101)

Work is not force times distance. It is the component of the force ALONG the displacement, times the distance: $W = F d\cos\theta$, where $\theta$ is the angle between the force vector and the displacement vector. The formula sheet writes the same line as a dot product, $W = \vec{F}\cdot\vec{d}$, and the dot product $\vec{a}\cdot\vec{b} = ab\cos\theta$ of its vectors line is exactly $Fd\cos\theta$. Here $\theta = 25^{\circ}$, so $W = 60 \times 12.0 \times \cos 25^{\circ} = 652.5$ J. Do the check in two steps the first few times: the horizontal component of the pull is $60\cos 25^{\circ} = 54.4$ N, and $54.4 \times 12.0 = 652.5$ J. The vertical component, $60\sin 25^{\circ} = 25.4$ N, travels no distance, so it contributes nothing. Writing $W = 60 \times 12.0 = 720$ J costs the whole mark, and it is the single most common line on a first energy test.

Gravity points straight down and the displacement is horizontal, so the angle between them is $90^{\circ}$ and $\cos 90^{\circ} = 0$: the weight does ZERO work on this trip. The normal force points straight up, same angle, same conclusion. This is not a special case, it is the rule that makes energy methods usable: a force perpendicular to the motion never appears in an energy balance. The normal force on a slope, the tension in a pendulum string and the string tension in circular motion all drop out for exactly this reason. Note what the sentence does NOT say: gravity and the normal force are not zero, and they are not useless. They cancel vertically and keep the cart on the floor. They simply transfer no energy.

Friction acts backwards along the motion, so $\theta = 180^{\circ}$ and $\cos 180^{\circ} = -1$: $W_{f} = 22 \times 12.0 \times (-1) = -264$ J. The minus sign is the physics, not a decoration. It says that 264 J left the cart and went into heating the floor and the wheels. A positive 264 J written here would mean friction SPEEDS the cart up, and the rest of the question would then be wrong by 528 J.

Add the works, do not add the forces at different angles: $W_{\text{net}} = 652.5 - 264 = 388.5$ J, the two zero terms changing nothing. The work-energy theorem gives $W_{\text{net}} = \frac{1}{2}mv^{2} - 0$, so $v = \sqrt{\dfrac{2 \times 388.5}{18}} = 6.57$ m/s. Notice what was never needed: the time, the acceleration, and the mass of anything but the cart. That is the whole argument for energy methods, and it is why this question would be painful with kinematics alone.

Hooke's law says $F = kx$, so the graph is a straight line through the origin and $k$ is its SLOPE, in newtons per metre: $k = \dfrac{24}{0.40} = 60$ N/m. Reading a single pair of values off the line is enough because the line passes through the origin; on a line that did not, you would need two points and a difference.

The area under the line from $0$ to $0.40$ m is a triangle of base $0.40$ m and height 24 N: $W = \frac{1}{2} \times 0.40 \times 24 = 4.8$ J. This is where the formula $W = \frac{1}{2}kx^{2}$ comes from, and you can confirm it: $\frac{1}{2} \times 60 \times 0.40^{2} = 4.8$ J. Use the geometry when a graph is given and the formula when $k$ is given; they are the same statement. A very common wrong line is $W = Fx = 24 \times 0.40 = 9.6$ J, exactly twice too much, because it treats the force as if it had been 24 N the whole way. It was 24 N only at the very end, and 0 N at the start.

Two clean ways. Subtract areas: the triangle to $0.40$ m is $4.8$ J, the triangle to $0.20$ m is $\frac{1}{2} \times 0.20 \times 12 = 1.2$ J, so the strip between them is $4.8 - 1.2 = 3.6$ J. Or read the strip directly as a trapezoid of parallel sides 12 N and 24 N and width $0.20$ m: $\frac{1}{2}(12 + 24) \times 0.20 = 3.6$ J. The two agree, which is the check.

The first half of the stretch costs $1.2$ J and the second half costs $3.6$ J, a ratio of 3, not 1. The reason is that elastic energy goes as the SQUARE of the extension: doubling $x$ multiplies the stored energy by 4, so the second half must carry the missing three quarters. Expect this everywhere in this chapter. A car at twice the speed has four times the kinetic energy, a spring at twice the extension has four times the stored energy, and anything that reasons 'twice as far, so twice the energy' is wrong by a factor that grows.

Two states, both momentarily at rest: the release, and the instant of greatest compression. Zero of gravitational potential energy: the position of the parcel at greatest compression. The system is the parcel, the buffer spring and the Earth; friction is the only non-conservative force.

Between the two states the parcel slides $D + x$ along the chute: the $D$ before it touches the buffer AND the $x$ while it squeezes it. Gravity and friction keep working over that last stretch.

On the chute nothing accelerates perpendicular to the surface, so $F_N = mg\cos\theta$ and the friction force is $F_{fr} = \mu_k mg\cos\theta$, pointing UP the chute, against the motion.

Extended work-energy principle, $W_{nc} = \Delta KE + \Delta PE$:

$\frac{1}{2}kx^{2} = F(D + x)$, with $F = mg(\sin\theta - \mu_k\cos\theta)$, the net pull along the chute while the parcel slides.

This is the quadratic $\frac{1}{2}kx^{2} - Fx - FD = 0$, solved with the formula of the sheet; the root with the minus sign is negative, a stretched spring, and is rejected:

Unit: $F/k$ is N divided by N/m, a length; $F^{2}$ and $kFD$ are both in N$^{2}$, so the square root is in newtons and $x$ is in metres.

Case $D = 0$, the parcel let go touching the buffer: $x = \dfrac{2F}{k}$, twice the compression $F/k$ at which the spring would simply hold the parcel. Released against the buffer, the parcel overshoots that balance point by as much again.

A stickier chute: a larger $\mu_k$ makes $F$ smaller, and $x$ grows with $F$, so $x$ shrinks. At $\mu_k = \tan\theta$, $F = 0$ and $x = 0$: the parcel never starts to slide. An $x$ that grew with $\mu_k$ would carry the friction work with the wrong sign.

Pitfall: stopping the slide at the free end of the buffer, $\frac{1}{2}kx^{2} = FD$, forgets that gravity and friction still act during the compression; it gives an $x$ that is too small.

The net pull along the chute:

$F = 8.0 \times 9.81 \times (\sin 30^{\circ} - 0.20\cos 30^{\circ}) = 25.6$ N

$x = \dfrac{25.6 + \sqrt{25.6^{2} + 2 \times 900 \times 25.6 \times 1.50}}{900} = \dfrac{25.6 + 264.4}{900} = 0.322$ m

Check in the balance itself: $\frac{1}{2} \times 900 \times 0.322^{2} = 46.7$ J, and $F(D + x) = 25.6 \times 1.822 = 46.7$ J.

The version that forgets the compression stretch, $x = \sqrt{2FD/k}$, gives $0.292$ m, about nine per cent short.

Thermal energy: $\mu_k mg\cos\theta\,(D + x) = 13.6 \times 1.822 = 24.8$ J.

Energy check: gravity gives up $mg(D + x)\sin\theta = 71.5$ J, which is $46.7$ J stored in the buffer plus $24.8$ J of heat.

What happens next is a question about forces: at full compression the buffer pushes with $kx = 290$ N, far more than the chute can hold back, so the parcel is thrown back up. A real buffer carries a damper for that reason.

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2} \times 1400 \times 25.0^{2} = 437\,500$ J, that is $4.375 \times 10^{5}$ J. Square the speed BEFORE multiplying: $\frac{1}{2} \times 1400 \times 25.0 = 17\,500$ is a wrong line that looks plausible and is 25 times too small.

The braking force is the only horizontal force, it points opposite the motion, and the car ends at rest. The work-energy theorem gives $-Fd = 0 - E_{k}$, so $F = \dfrac{437\,500}{48.0} = 9114.6 \approx 9.11 \times 10^{3}$ N. Sanity check: the weight of the car is $1400 \times 9.8 = 13\,720$ N, so the braking force is about two thirds of the weight, which is a hard but realistic stop on dry asphalt. A force of $10^{5}$ N or of 900 N would both be signals to go back.

Same force, new energy: $E_{k} = \frac{1}{2} \times 1400 \times 50.0^{2} = 1\,750\,000$ J, and $d = \dfrac{1\,750\,000}{9114.6} = 192$ m. Four times the distance, from twice the speed, with no new physics.

Because the quantity the brakes have to remove is the kinetic energy, and it grows as the SQUARE of the speed. Twice the speed means four times the energy; with the same force acting, $d = E_{k}/F$ gives four times the distance. This is also the answer to the road-safety version of the question: between 50 km/h and 100 km/h, the stopping distance does not double, it quadruples.

$U = mg\Delta y = 0.150 \times 9.8 \times 1.20 = 1.764$ J, where $\Delta y$ is measured from the chosen zero.

From the floor the ball is $1.20 + 0.80 = 2.00$ m up, so $U = 0.150 \times 9.8 \times 2.00 = 2.94$ J. Same ball, same instant, two different numbers. Neither is wrong, because potential energy is not a property of the object alone; it is a bookkeeping device that only ever appears in a DIFFERENCE.

Table zero: the ball goes from $U = 1.764$ J to $U = 0$, so $\Delta U = -1.764$ J. Floor zero: from $2.94$ J to $0.150 \times 9.8 \times 0.80 = 1.176$ J, so $\Delta U = 1.176 - 2.94 = -1.764$ J. Identical, as it must be. With no friction, $\frac{1}{2}mv^{2} = 1.764$ J gives $v = \sqrt{\dfrac{2 \times 1.764}{0.150}} = 4.85$ m/s, and the mass cancels if you write it as $v = \sqrt{2g h} = \sqrt{2 \times 9.8 \times 1.20} = 4.85$ m/s. The speed does not depend on the zero level, which is the point of the whole exercise.

The sentence is missing the reference level. 'The ball has $2.94$ J of gravitational potential energy RELATIVE TO THE FLOOR' is a complete statement; without the last three words the number could equally have been $1.764$ J. On an exam this costs marks twice over: once for the incomplete statement, and again whenever a student uses one zero in the first line and the other in the second, which produces a bogus $1.176$ J of energy out of nowhere. Write the zero level in words before any numbers, for example: taking the table top as the zero of gravitational potential energy.

The bob hangs $L\cos\theta$ below the pivot when the string makes an angle $\theta$, and $L$ below it at the bottom. The rise is therefore $h = L - L\cos\theta = L(1 - \cos\theta) = 1.60\,(1 - \cos 35^{\circ}) = 1.60 \times 0.1808 = 0.289$ m. Two wrong heights are collected every year on this one: $L\sin\theta = 0.918$ m, which is the sideways displacement and not a height at all, and $L\cos\theta = 1.31$ m, which is the depth below the pivot rather than the rise above the bottom. Both give a bob moving far too fast.

The string does no work (part c), the only other force is gravity, so the mechanical energy is conserved: $mgh = \frac{1}{2}mv^{2}$, the mass cancels, and $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.289} = 2.38$ m/s. The fact that the mass cancels is worth noticing: a bowling ball and a marble on the same string reach the bottom at the same speed, and any answer that depends on the $0.40$ kg has a mistake in it.

Zero. The tension always points along the string, towards the pivot, while the velocity is always along the arc, perpendicular to the string. The angle between force and displacement stays $90^{\circ}$ for the whole swing, and $\cos 90^{\circ} = 0$. This is why the pendulum can be solved with energy at all: the only force left in the balance is gravity, whose work depends on nothing but the change in height.

Half the maximum speed is $v' = 1.191$ m/s, and the height at which that happens follows from $\frac{1}{2}mv'^{2} = mgh'$, so $h' = \dfrac{v'^{2}}{2g} = 0.0723$ m. Then $h' = L(1 - \cos\theta)$ gives $\cos\theta = 1 - \dfrac{0.0723}{1.60} = 0.9548$ and $\theta = 17.3^{\circ}$. Half the speed is a QUARTER of the height, not half of it, and it is certainly not half the angle: $17.5^{\circ}$ would be the answer to a question about angles, and this one is about energy.

Take B as the zero of gravitational potential energy and say so on the copy. With no friction, $mgh = \frac{1}{2}mv^{2}$, the mass cancels, and $v = \sqrt{2 \times 9.8 \times 28.0} = 23.4$ m/s. The shape of the hill, its steepness and its length are all irrelevant: gravity is a conservative force and its work depends only on the 28.0 m of vertical drop.

The energy available at A is $E = mgh = 320 \times 9.8 \times 28.0 = 87\,808$ J. What actually arrived at B is $\frac{1}{2} \times 320 \times 21.0^{2} = 70\,560$ J. The difference, $87\,808 - 70\,560 = 17\,248$ J, went into the wheels, the rails and the air, which is $17\,248/87\,808 = 0.196$, about 19.6 per cent. Say it correctly: the mechanical energy was not conserved, but no energy was destroyed. It became thermal energy, and the total is still $87\,808$ J.

One line does the whole thing, and it is the line to memorise: energy at the start, plus the work of the non-conservative forces, equals energy at the end. From B to C, $70\,560 - 12\,000 = \frac{1}{2}(320)v^{2} + 320 \times 9.8 \times 18.0$. The potential term is $56\,448$ J, leaving $70\,560 - 12\,000 - 56\,448 = 2112$ J of kinetic energy, so $v = \sqrt{\dfrac{2 \times 2112}{320}} = 3.63$ m/s. A small number, and it should be: the car barely crests the hill, which is exactly what a designer aims for.

The car stops climbing when all the kinetic energy it still has is gone. Starting from the $70\,560$ J at B and paying the same $12\,000$ J, the ceiling is $mgh = 70\,560 - 12\,000 = 58\,560$ J, so $h = \dfrac{58\,560}{320 \times 9.8} = 18.7$ m. The 18.0 m hill is cleared, with about $0.7$ m to spare, and that is the margin an engineer would call thin. Note what this part did NOT need: any knowledge of the shape of the track between B and C, and any time.

Only the VERTICAL rise counts, so the length of each step and the number of steps do not enter: $W = mgh = 68 \times 9.8 \times 4.2 = 2799$ J. Average power is work divided by time: $P = \dfrac{2799}{6.5} = 431$ W. This is a believable figure for a fit student sprinting up stairs, roughly half a horsepower, and it can be sustained for seconds rather than minutes. If your answer came out at 4 W or at 40 kW, the time or the height went in wrong.

At constant speed the motor's force exactly balances the weight, $F = mg = 11\,760$ N, so $P = Fv = 11\,760 \times 2.5 = 29\,400$ W, that is $2.94 \times 10^{4}$ W. The identity $P = Fv$ is simply $W/t$ with $W = Fd$ and $d/t = v$, and it is the fastest route whenever a problem gives a constant speed. Writing $P = mgh$ with some height would be a category error: that product is an energy in joules, not a power in watts.

Efficiency is useful output over total input: $\dfrac{29\,400}{35\,000} = 0.84$, so 84 per cent. The missing 16 per cent, $5600$ W, heats the motor windings and the cables. Efficiency has no units and can never exceed 1; an answer above 100 per cent means the input and output were swapped.

At constant speed the cyclist's driving force equals the resistance, 32 N, so $P = Fv = 32 \times 9.0 = 288$ W. Nothing is accelerating and nothing is climbing, so every joule goes straight into air drag and rolling resistance. Compare with part a): the same person can produce 431 W for a few seconds on a staircase and about 288 W for an hour on a bicycle, which is what average power really measures, and why a cyclist's figure is always quoted with a duration attached.

The acceleration is constant because the force is: $a = \dfrac{24.0}{3.0} = 8.0$ m/s$^{2}$.

$F = ma = 6.0 \times 10^{3} \times 8.0 = 4.8 \times 10^{4}$ N

At the end of the launch: $P = Fv = 4.8 \times 10^{4} \times 24.0 = 1.152 \times 10^{6}$ W, about $1.15$ MW (formula sheet, power line: $P = \vec{F}\cdot\vec{v}$).

The force is fixed and $v = at$ grows from zero, so $P = Fat$ is proportional to time: half the final power when the speed is $12.0$ m/s, at $t = 1.5$ s, half-way through the launch.

Energy given to the train: $KE = \frac{1}{2} \times 6.0 \times 10^{3} \times 24.0^{2} = 1.728 \times 10^{6}$ J.

Area check: the power rises in a straight line from $0$ to $1.152$ MW over $3.0$ s, a triangle of area $\frac{1}{2} \times 1.152 \times 10^{6} \times 3.0 = 1.728 \times 10^{6}$ J. The same number with no calculus: the area under a power graph is an energy.

Design consequence: the average power is $\dfrac{1.728 \times 10^{6}}{3.0} = 5.76 \times 10^{5}$ W, only half the peak, yet the supply must be built for the peak.

Pitfall: $P = F \times$ the average speed gives the AVERAGE power, $4.8 \times 10^{4} \times 12.0 = 5.76 \times 10^{5}$ W, not the power at the end.

Graph B, the straight line that starts at zero.

The force is the same at every instant, and $P = Fv$ with $v = at$: the power is proportional to time, zero at the start and largest at the end.

Graph A, a constant power, would need a force that DROPS as the speed rises, $F = P/v$: that is a motor working flat out, not a motor pushing with a fixed force.

Graph C, a curve bending upwards, is the kinetic energy given so far, $\frac{1}{2}m(at)^{2}$, which grows as the square of time: an energy, not a rate. It is the area under B.

Graph D, a power that falls, is the intuition that the motors work hardest when the train is hardest to set moving. At the start the force is full but the speed is zero, so no energy is being transferred yet.

Pitfall: confusing a large force with a large power. Power needs both a force and a speed along it.

FALSE. Work is $W = Fd\cos\theta$, and a force at $90^{\circ}$ to the displacement gives $\cos 90^{\circ} = 0$ no matter how large it is. The normal force on a sledge, the tension in a pendulum string and the centripetal force on a satellite in a circular orbit are all examples: they change the direction of the motion without changing the speed. Correct sentence: a force does work only if it has a component along the displacement.

FALSE, and this is the single most expensive line in the chapter. $E_{k} = \frac{1}{2}mv^{2}$, so doubling $v$ multiplies the energy by $2^{2} = 4$. A 1200 kg car at 15 m/s carries $135\,000$ J; at 30 m/s it carries $540\,000$ J, not $270\,000$ J. Correct sentence: twice the speed means four times the kinetic energy, and four times the stopping distance for the same braking force.

FALSE. Gravity is a conservative force: its work is $-mg\Delta y$ and depends only on the change in HEIGHT between the start and the end, not on the path. A hiker who climbs 800 m by a short steep trail or by a long gentle one does the same work against gravity. What the two trails do change is the friction, the duration and the power, not the work of gravity. Correct sentence: the work of gravity depends only on the vertical drop or rise.

FALSE as written, and the mistake is in the last three words. Energy is never destroyed. When friction acts, the MECHANICAL energy $E_{k} + U$ decreases, and exactly the same amount appears as thermal energy in the surfaces and the air. Correct sentence: friction converts mechanical energy into thermal energy, so the mechanical energy decreases while the total energy stays constant.

TRUE. $U = mg\Delta y$ is measured from a level YOU choose, so an object below that level has a negative $\Delta y$ and a negative $U$. A diver 3 m below the surface has $U = -3mg$ if the surface is the zero. Nothing is wrong with that, because only DIFFERENCES in potential energy appear in any equation you will write. What would be impossible is a negative kinetic energy, since $v^{2} \ge 0$ always.

Convert the time FIRST: 22 minutes is $22 \times 60 = 1320$ s. Then $W = Pt = 95 \times 1320 = 125\,400$ J, that is $1.254 \times 10^{5}$ J. Leaving the time in minutes gives 2090 J, sixty times too small, and it is the most frequent error in the whole power section; a watt is a joule per SECOND, so any time that enters a power formula is in seconds.

Efficiency is useful output over input, so the input is the output divided by the efficiency: $E = \dfrac{125\,400}{0.25} = 501\,600$ J. Dividing by 4184 J per Calorie gives $\dfrac{501\,600}{4184} = 120$ Calories. Multiplying by $0.25$ instead of dividing is the classic slip and gives $31\,350$ J, less than the work actually delivered, which is physically impossible and should be caught on sight.

Whatever is not mechanical work is heat: $501\,600 - 125\,400 = 376\,200$ J, three times the useful work. This is why a spinning class heats a room and why the patient sweats: three quarters of the chemical energy consumed never reaches the pedals.

The metabolic power is $\dfrac{95}{0.25} = 380$ W, and the bar holds $230 \times 4184 = 962\,320$ J of chemical energy. So $t = \dfrac{962\,320}{380} = 2532$ s, that is 42 minutes. The number is worth remembering for the next time a label is read as a promise: a single chocolate bar is roughly three quarters of an hour of steady effort, and the 'calories burned' display on a machine reports the chemical input, not the mechanical work, which is why it reads four times higher than the work at the pedals.

During the fall only gravity does work, so $mgh = \frac{1}{2}mv^{2}$ and the mass cancels: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.90} = 4.2$ m/s. Note that this is a fall from a little under a metre, a height nobody thinks twice about.

$E_{k} = \frac{1}{2} \times 72 \times 4.2^{2} = 635$ J. Equivalently $E_{k} = mgh = 72 \times 9.8 \times 0.90 = 635$ J, which is the faster route and a useful check on part a).

During the landing the body travels a further distance $d$ downwards while gravity still pushes down and the ground pushes up with an average force $F$. The work-energy theorem over the landing gives $(mg - F)d = 0 - E_{k}$, so $F = \dfrac{E_{k}}{d} + mg = \dfrac{635.04}{0.010} + 705.6 = 64\,210$ N, that is $6.42 \times 10^{4}$ N. The weight term is only $705.6$ N here, about 1 per cent, but writing it shows the marker that the reasoning is complete.

With $d = 0.45$ m, $F = \dfrac{635.04}{0.45} + 705.6 = 2117$ N, that is $2.12 \times 10^{3}$ N. The person's weight is $72 \times 9.8 = 705.6$ N, so the stiff landing loads the body at 91 times its own weight and the bent-knee landing at 3 times. The two forces are in the ratio $\dfrac{64\,210}{2117} = 30.3$, not 45, the ratio of the two stopping distances: only the part $E_{k}/d$ scales as $1/d$ ($63\,504$ N against $1411$ N, exactly 45 times), while the weight term, $705.6$ N, is the same in both landings and pulls the ratio down. Nothing about the fall changed: the same 635 J had to go somewhere, and the only variable was the DISTANCE over which it was removed. Every protective device works on this single line, $F = E_{k}/d$: a crash helmet, a gymnastics mat, a climbing rope that stretches, an airbag. They do not remove energy, they lengthen $d$.