Where Did the Missing Joules Go?
A $5\ \text{kg}$ block slides $12\ \text{m}$ down a chute of vertical drop $6\ \text{m}$. On the smooth chute it would arrive at speed $10.8\ \text{m s}^{-1}$; in practice it arrives slower. Set the friction force with the slider and watch the ledger: the potential energy released splits into the kinetic energy delivered and the work handed to friction. Nothing has vanished — the audit has a line for everything.
energy released $mgh = 294$ J · kinetic energy at the bottom $mgh - R \times 12$ = 174 J · speed 8.34 m s⁻¹
The bar is the whole $294\ \text{J}$: the green part is what the block still carries at the bottom, the amber part is what friction has taken. When the amber part grows, less reaches the bottom.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. A block slides down a smooth slope of height 6 m from rest. Its speed at the bottom is about:
R2. A crate is dragged 8 m across a floor against a friction force of 50 N. The work done against friction is:
R3. A car has a driving force of 600 N and faces a resistance of 600 N. It is:
Energy is never lost, but friction and drag transfer it away from the mechanical account as heat. For a descent down a rough surface:
$mgh = \dfrac{1}{2}mv^{2} + R \times L$
where $L$ is the length of the path — not the height dropped. Mini-example: a $3\ \text{kg}$ block descends $4\ \text{m}$ along a $10\ \text{m}$ path against $5\ \text{N}$: released $117.6\ \text{J}$, friction takes $50\ \text{J}$, so the block carries $67.6\ \text{J}$ and leaves at $6.71\ \text{m s}^{-1}$.
If an audit shows that energy $E_{removed}$ left the account over a path of length $L$, the average resistance force is
$R = \dfrac{E_{removed}}{L}$
Mini-example: a car of mass $800\ \text{kg}$ braking from $20\ \text{m s}^{-1}$ to $10\ \text{m s}^{-1}$ in $50\ \text{m}$ removes $\dfrac{1}{2} \times 800 \times (20^{2} - 10^{2}) = 120\,000\ \text{J}$, so the average braking force is $\dfrac{120\,000}{50} = 2400\ \text{N}$.
Write the account for the whole journey: engine work and any energy released by falling on one side; kinetic energy gained, potential energy stored and energy removed by resistance on the other.
A lorry of mass $10\,000\ \text{kg}$ accelerates from $10\ \text{m s}^{-1}$ to $20\ \text{m s}^{-1}$ while climbing $20\ \text{m}$, against a resistance of $3000\ \text{N}$ over $200\ \text{m}$. The engine must supply $\dfrac{1}{2} \times 10\,000 \times (400 - 100) = 1.5 \times 10^{6}\ \text{J}$ of kinetic energy, $10\,000 \times 9.8 \times 20 = 1.96 \times 10^{6}\ \text{J}$ of potential energy and $3000 \times 200 = 6 \times 10^{5}\ \text{J}$ against resistance: $4.06 \times 10^{6}\ \text{J}$ in total.
Groups of 4: run the same 3 kg block down two chutes of identical geometry — one rough, one smooth — with the bars live. Watch GPE empty, KE fill, and on the rough chute the heat bar claim the difference. Reproduce the audit on paper as an equation, then settle the three graded challenges.
Work alone on two resistance laws: a constant rough patch and a resistance that grows with distance. Predict the readout before you scrub the slider. The growing case is the one ordinary force-times-distance cannot handle — the shaded area must price the work. Then take the three graded challenges.
The forces acting on a hot air balloon of mass $500\ \text{kg}$ are its weight and the total uplift force. (i) Find the total work done when the speed of the balloon changes from (a) $2\ \text{m s}^{-1}$ to $5\ \text{m s}^{-1}$ (b) $8\ \text{m s}^{-1}$ to $3\ \text{m s}^{-1}$. (ii) If the balloon rises $100\ \text{m}$ vertically while its speed changes, calculate in each case the work done by the uplift force.
"Total work" means the work of the resultant — which equals the change in kinetic energy. For the uplifting force in part (ii), the audit adds the potential energy gained: uplift work = kinetic change + $mgh$. Watch the signs — losing speed makes one of the contributions negative.
(a) $\dfrac{1}{2} \times 500 \times (5^{2} - 2^{2}) = 250 \times 21 = 5250\ \text{J}$. (b) $\dfrac{1}{2} \times 500 \times (3^{2} - 8^{2}) = 250 \times (-55) = -13\,750\ \text{J}$.
The rise stores $mgh = 500 \times 9.8 \times 100 = 490\,000\ \text{J}$ of potential energy, and the uplift force must fund both that and the kinetic change: (a) $5250 + 490\,000 = 495\,250\ \text{J}$; (b) $-13\,750 + 490\,000 = 476\,250\ \text{J}$.
In (b) part of the balloon's kinetic energy helps pay for the climb, exactly as the ledger demands: even though gravity alone would need $490\ \text{kJ}$, the lift supplies slightly less because the balloon is slowing down.
A lorry of mass $12\,500\ \text{kg}$ travels along a road with a straight horizontal section $AB$ and a straight inclined section $BC$ of length $500\ \text{m}$. The speeds of the lorry at $A$, $B$ and $C$ are $17\ \text{m s}^{-1}$, $25\ \text{m s}^{-1}$ and $17\ \text{m s}^{-1}$. (i) The work done against the resistance to motion from $A$ to $B$ is $5000\ \text{kJ}$. Find the work done by the driving force as the lorry travels from $A$ to $B$. (ii) As the lorry travels from $B$ to $C$, the resistance is $4800\ \text{N}$ and the work done by the driving force is $3300\ \text{kJ}$. Find the height of $C$ above the level of $AB$.
Write the audit for each section. On $AB$ the lorry gains kinetic energy, so the driving force must cover that plus the resistance work. On $BC$ it gains height but loses speed — the speed loss helps pay for the climb.
Gain in kinetic energy $= \dfrac{1}{2} \times 12\,500 \times (25^{2} - 17^{2}) = 6250 \times 336 = 2\,100\,000\ \text{J} = 2100\ \text{kJ}$. Driving work $= 2100 + 5000 = 7100\ \text{kJ}$.
Resistance work $= 4800 \times 500 = 2\,400\,000\ \text{J} = 2400\ \text{kJ}$; kinetic change $= \dfrac{1}{2} \times 12\,500 \times (17^{2} - 25^{2}) = -2100\ \text{kJ}$. Audit: $3300 - 2400 - mgh = -2100$ (all in kJ), so $mgh = 3000\ \text{kJ}$.
$h = \dfrac{3\,000\,000}{12\,500 \times 9.8} = 24.5\ \text{m}$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. A 2 kg block slides down a smooth slope from a height of 3 m. Its kinetic energy at the bottom is:
Q2. A 3 kg block slides down a rough slope of height 4 m (path length 10 m) and arrives with 60 J of kinetic energy. The energy removed by friction is:
Q3. A car of mass 1000 kg slows from 20 m/s to 10 m/s over 75 m. The average resistance is:
Q4. A 400 kg balloon rises 50 m while its speed rises from 4 m/s to 6 m/s. The work done by the uplift force is:
Work each on paper, then reveal the solution to check your working.
A car of mass $1600\ \text{kg}$ is travelling at $25\ \text{m s}^{-1}$ when the brakes are applied so that it stops after moving a further $75\ \text{m}$. (i) Find the work done by the brakes. (ii) Find the retarding force from the brakes, assuming it is constant and that other resistive forces may be neglected.
Reveal Solution
(i) The brakes must remove all the kinetic energy: $\dfrac{1}{2} \times 1600 \times 25^{2} = 500\,000\ \text{J} = 500\ \text{kJ}$. (ii) $F = \dfrac{500\,000}{75} = 6667\ \text{N}$ (about $6.7\ \text{kN}$).
A ski-run starts at altitude $2471\ \text{m}$ and ends at $1863\ \text{m}$. (i) If all resistance forces could be ignored, what would be the speed of the skier at the end of the run? A particular skier of mass $70\ \text{kg}$ actually attains a speed of $42\ \text{m s}^{-1}$. The length of the run is $3.1\ \text{km}$. (ii) Find the average force of resistance acting on the skier. (iii) Two skiers are equally skilful — which would you expect to be travelling faster by the end of the run, the heavier or the lighter?
Reveal Solution
The drop is $2471 - 1863 = 608\ \text{m}$. (i) $v = \sqrt{2 \times 9.8 \times 608} = 109\ \text{m s}^{-1}$ (3 s.f.). (ii) Energy removed $= 70 \times 9.8 \times 608 - \dfrac{1}{2} \times 70 \times 42^{2} = 417\,100 - 61\,740 = 355\,400\ \text{J}$, over $3100\ \text{m}$: $R = \dfrac{355\,400}{3100} = 115\ \text{N}$ (3 s.f.). (iii) The heavier skier: both release energy in proportion to their mass, but the resistance forces grow much more slowly, so the heavier skier keeps a larger share and finishes faster.
Akosua draws water from a well $12\ \text{m}$ below the ground. Her bucket holds $5\ \text{kg}$ of water and by the time she has pulled it to the top of the well it is travelling at $1.2\ \text{m s}^{-1}$. (i) How much work does Akosua do in drawing the bucket of water? On an average day 150 people in the village each draw six such buckets of water. One day a new electric pump is installed that takes water from the well and fills an overhead tank $5\ \text{m}$ above ground level every morning, the water having speed $2\ \text{m s}^{-1}$ on arriving in the tank. (ii) How much work does the pump do in one day? (iii) It takes the pump one hour to fill the tank each morning — at what rate does it do work, in joules per second?
Reveal Solution
(i) $5 \times 9.8 \times 12 + \dfrac{1}{2} \times 5 \times 1.2^{2} = 588 + 3.6 = 592\ \text{J}$. (ii) Water per day $= 150 \times 6 \times 5 = 4500\ \text{kg}$, raised $17\ \text{m}$: $4500 \times 9.8 \times 17 + \dfrac{1}{2} \times 4500 \times 2^{2} = 749\,700 + 9000 = 758\,700\ \text{J}$. (iii) $\dfrac{758\,700}{3600} = 211\ \text{W}$.
Wesley, whose mass is $70\ \text{kg}$, inadvertently steps off a bridge $50\ \text{m}$ above water. When he hits the water, Wesley is travelling at $25\ \text{m s}^{-1}$. (i) Calculate the potential energy Wesley has lost and the kinetic energy he has gained. (ii) Find the size of the resistance force acting on him while he is in the air, assuming it to be constant. Wesley descends to a depth of $5\ \text{m}$ below the water surface, then returns to the surface. (iii) Find the total upthrust (assumed constant) acting on him while he is moving downwards in the water.
Reveal Solution
(i) Potential energy lost $= 70 \times 9.8 \times 50 = 34\,300\ \text{J}$; kinetic energy gained $= \dfrac{1}{2} \times 70 \times 25^{2} = 21\,875\ \text{J}$. (ii) The air resistance removed the difference over $50\ \text{m}$: $R = \dfrac{34\,300 - 21\,875}{50} = 249\ \text{N}$. (iii) In the water he descends $5\ \text{m}$ and stops. The upthrust must remove his remaining kinetic energy and overcome gravity on the way down: $U \times 5 = 21\,875 + 70 \times 9.8 \times 5 = 21\,875 + 3430 = 25\,305$, so $U = 5061\ \text{N}$ (about $5.1\ \text{kN}$). This upthrust exceeds his weight, which is why he then rises back to the surface.
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