Exercises — Power — average and instantaneous, units

WHAT tool: "work done in a time" → average power, $P_{\text{avg}}=W/\Delta t$. The word average and a whole interval (not an instant) is the giveaway. $P_{\text{avg}}=\frac{6000}{4}=1500\text{W}=1.5\text{kW}.$

WHAT tool: rearrange $P=W/\Delta t$ into $W=P\Delta t$ because we now know power and want energy. $1$ minute $=60$ s. $W=P\Delta t=60\times 60=3600\text{J}.$ A $60$ W bulb burns $60$ J every second — over $60$ s that's $3600$ J.

WHAT tool: we want power right now with a force and a speed, so $P=Fv\cos \theta$. Same direction means the angle between $\vec{F}$ and $\vec{v}$ is $\theta =0^{\circ }$, and $\cos 0^{\circ }=1$. $P=Fv\cos 0^{\circ }=2000\times 15\times 1=30000\text{W}=30\text{kW}.$

Step 1 — WHAT and WHY: at constant speed the lifting force just balances weight, so the work done is the work against gravity, $W=mgh$. (Uses the ideas from Work — definition and W = F·d cosθ.) $W=mgh=300\times 10\times 20=60000\text{J}.$ Step 2: rate this over the time. $P_{\text{avg}}=\frac{W}{\Delta t}=\frac{60000}{50}=1200\text{W}=1.2\text{kW}.$

WHAT tool: $P=Fv\cos \theta$, because $\theta =60^{\circ }$ is the angle between the tilted push $\vec{F}$ and the horizontal velocity $\vec{v}$. This is exactly the dot product $\vec{F}\cdot \vec{v}$ — it keeps only the component of $\vec{F}$ pointing along $\vec{v}$. $P=Fv\cos 60^{\circ }=50\times 2\times \frac{1}{2}=50\text{W}.$ The vertical component of the push ($F\sin 60^{\circ }$) is perpendicular to the motion, so it delivers zero power — that's the whole reason $\cos \theta$ and not $\sin \theta$ appears.

Step 1 — WHY use kinetic energy: the work done equals the change in kinetic energy (the work–energy theorem), because all the work went into speeding the car up. $W=\Delta KE=\frac{1}{2}m{v}^2-0=\frac{1}{2}(1000)(20{)}^2=200000\text{J}.$ Step 2: $P_{\text{avg}}=\frac{W}{\Delta t}=\frac{200000}{8}=25000\text{W}=25\text{kW}.$

Force (constant): $F=ma=2\times 4=8\text{N}$. Velocity at $t=3$: $v=at=4\times 3=12\text{m/s}$.

(a) Instantaneous power at $t=3$: $P_{\text{inst}}(3)=Fv=8\times 12=96\text{W}.$ WHY it grows: the force is fixed, but $v$ increases with time, so $P=Fv$ increases. In fact $P(t)=F(at)=8\times 4t=32t$ — a straight line through the origin (red line in the figure).

(b) Average power: total work via kinetic energy, $W=\frac{1}{2}m{v}^2=\frac{1}{2}(2)(12{)}^2=144\text{J}$. $P_{\text{avg}}=\frac{144}{3}=48\text{W}.$ Notice $48=\frac{1}{2}\times 96$: since power rose linearly from $0$ to $96$, its average is exactly the midpoint (the shaded triangle's area ÷ base in the figure).

WHY integrate: power is $dW/dt$, the rate of doing work. To recover total work from a varying rate we must add up $Pdt$ over the interval — that is an integral (equivalently, the area under the $P$–$t$ graph). $W={\int }_0^{5}Pdt={\int }_0^{5}6tdt={\left[3t^2\right]}_0^5=3(25)=75\text{J}.$ Geometric check: the graph of $P=6t$ is a triangle of base $5$ and height $6\times 5=30$; area $=\frac{1}{2}(5)(30)=75$ J. ✓

At steady speed the drive force equals the resistance, so $P=Fv$. $P_1=30\times 8=240\text{W},P_2=30\times 16=480\text{W}.$ $\frac{P_2}{P_1}=\frac{480}{240}=2.$ With force held fixed, power scales linearly with speed — double the speed, double the power.

Step 1 — useful output power: the useful job is lifting against gravity at constant speed, so the lifting force is the weight $mg$ and the useful power is $P_{\text{out}}=(mg)v$. (See Energy conservation and efficiency.) $P_{\text{out}}=mgv=20\times 10\times 2=400\text{W}.$ Step 2 — efficiency = useful out ÷ total in. $\eta =\frac{P_{\text{out}}}{P_{\text{in}}}=\frac{400}{500}=0.8=80\%.$ The missing $100$ W becomes heat and sound inside the motor.

WHY add two forces: at constant speed there's no kinetic-energy change, so the engine's drive force must balance both the component of gravity down the slope and the resistance.

• Gravity's pull along the slope: $mg\sin 30^{\circ }=1200\times 10\times 0.5=6000\text{N}$.
• Resistance: $500\text{N}$.

Total drive force $F=6000+500=6500\text{N}$, and this force is along the motion, so $\theta =0$: $P=Fv=6500\times 10=65000\text{W}=65\text{kW}.$

Step 1 — power arriving from the falling water. Each second, mass $\dot{m}=2000$ kg loses gravitational energy $mgh$. So the rate of energy arriving is $P_{\text{in}}=\dot{m}gh=2000\times 10\times 50=1000000\text{W}=1\text{MW}.$ WHY "per second" works here: power is energy per time, and $2000$ kg is the mass delivered each second, so $\dot{m}gh$ is already an energy-per-second, i.e. a power. Step 2 — apply efficiency: $P_{\text{out}}=0.9\times 1000000=900000\text{W}=0.9\text{MW}.$

WHY: at top speed the acceleration is zero, so the drive force equals the resistance, $F=600$ N. At maximum power the engine gives $P=75000$ W, and $P=Fv$, so $v_{\max }=\frac{P}{F}=\frac{75000}{600}=125\text{m/s}.$ Insight: top speed is set by power ÷ resistance, not by how big the force is alone. (In reality drag grows with speed, capping this far lower — but with the given constant resistance, $125$ m/s is the answer. Mass $1500$ kg is a distractor: it never enters a steady-speed power balance.)

(a) Energy in kWh — keep power in kW and time in hours so the units are kWh: $E=P\times t=3\text{kW}\times (5\times 30)\text{h}=3\times 150=450\text{kWh}.$ In joules, using $1\text{kWh}=3.6\times 10^6$ J: $E=450\times 3.6\times 10^6=1.62\times 10^9\text{J}.$ (b) Cost: $\text{cost}=450\times 8=₹3600.$

WHY calculus: we need $P=Fv$, and both $F$ and $v$ change with time, so we differentiate. Velocity is the rate of change of position (see Velocity and instantaneous rate (Kinematics)); acceleration is the rate of change of velocity.

• Velocity: $v=\frac{dx}{dt}=3t^2$. At $t=2$: $v=3(4)=12\text{m/s}$.
• Acceleration: $a=\frac{dv}{dt}=6t$. At $t=2$: $a=12{\text{m/s}}^2$.
• Force: $F=ma=2\times 12=24\text{N}$.

Force and velocity point the same way (both along $+x$), so $\theta =0$: $P=Fv=24\times 12=288\text{W}.$

Gravity's force is $\vec{F}=mg=0.5\times 10=5\text{N}$, pointing down. Velocity (up positive): $v(t)=20-10t$.

(a) At $t=1$ s: $v=20-10=10\text{m/s}$ (upward). Force is down, velocity up, so $\theta =180^{\circ }$, $\cos 180^{\circ }=-1$: $P=Fv\cos 180^{\circ }=5\times 10\times (-1)=-50\text{W}.$ Negative power: gravity is removing energy from the ball as it climbs.

(b) At $t=3$ s: $v=20-30=-10\text{m/s}$ (i.e. $10$ m/s downward). Now force and velocity both point down, $\theta =0$: $P=Fv\cos 0^{\circ }=5\times 10\times 1=+50\text{W}.$ Positive power: gravity is now giving energy back to the ball as it falls. The sign of power tells you the direction of energy flow.

• Energy from a $60$ W bulb in $60$ s? ::: $3600$ J.
• Push of $50$ N at $60^{\circ }$ to motion, speed $2$ m/s, power? ::: $50$ W.
• For constant force from rest, $P_{\text{avg}}$ vs $P_{\text{peak}}$? ::: $P_{\text{avg}}=\frac{1}{2}{P}_{\text{peak}}$.
• Sign of power when gravity acts on a rising ball? ::: Negative (energy removed).
• Top speed from power and resistance? ::: $v_{\max }=P/F_{\text{resist}}$.