IB Physics HLTopic 1 — Motion, Forces & EnergyPaper 1 & 2Work, Energy & Power~10 min read
Energy & Power
Two cars can reach the same speed, and two cranes can lift the same load to the same height — so they transfer the same energy. But if one does it in half the time, we say it’s more powerful. Power isn’t about how much energy is moved; it’s about how fast. That’s the whole idea here: power is the rate of doing work, and it gives us a second, very handy equation — power equals force times velocity — that turns up constantly in problems about vehicles, engines and machines.
📘 What you need to know
Power is the rate at which energy is transferred (or work is done)
P = ΔW / Δt — work done divided by time taken
For a constant force moving at constant velocity: P = Fv
Power is measured in watts (W), where 1 W = 1 J s−1
The same work done in less time means more power
In P = Fv, the force must be in the same direction as the velocity
Appliances have a power rating showing energy transferred per second
Power as the rate of doing work
Power tells you how quickly energy is being transferred. Since the energy transferred is the work done, power is the work done per unit time:
Power — rate of doing workP = ΔW / Δt
where P is the power in watts (W), ΔW is the work done (or energy transferred) in joules (J), and Δt is the time taken in seconds (s). Two machines might do the same total work, but the one that finishes sooner has developed more power.
Both motors lift the same weight through the same height h, doing equal work. The one that does it in less time (2 s vs 4 s) develops more power.
WE 1
A crane lifts a 250 kg load to a height of 12 m in 8.0 s. Calculate the average power developed by the crane. (Take g = 9.81 m s−2.)
Step 1 — the work done is the GPE gained
W = mgΔh
W = 250 × 9.81 × 12 = 29 430 JStep 2 — power is work ÷ time
P = ΔW ÷ Δt
Step 3 — substituteP = 29 430 ÷ 8.0P = 3700 W (2 s.f.)That’s about 3.7 kW — the rate at which the crane transfers energy to the load.
The power–velocity equation
There’s a second form of the power equation that’s often quicker. If a constant force F moves an object at a steady velocity v, then in a time Δt the object moves a distance vΔt, so the work done is FvΔt. Dividing by the time gives:
Power — force and velocityP = Fv
This form is perfect for vehicles cruising at steady speed, where an engine’s driving force balances the resistive forces. Note the two conditions: the force is constant and it points in the same direction as the velocity.
A vehicle driven forward by a constant force F at steady velocity v delivers power P = Fv. The force and velocity point the same way.
WE 2
A cyclist rides at a constant 6.5 m s−1 against a total resistive force of 22 N. Calculate the power the cyclist must develop to maintain this speed.
Step 1 — at constant speed, the driving force equals the resistance
F = 22 N
Step 2 — use P = FvP = 22 × 6.5P = 143 WAt steady speed there’s no acceleration, so the driving force just balances the 22 N of resistance.
Here’s the key link to remember: to keep something moving at a steady speed against resistance, the driving force only has to balance the resistive force — no more. So the power needed equals that resistive force times the speed. If a question gives you a drag force and a cruising speed, P = Fv is almost always the way in.
The watt
Power is measured in watts (W). One watt is a transfer of one joule of energy every second:
Definition of the watt
1 W = 1 J s−1
So a 1000 W (1 kW) appliance transfers 1000 joules every second. Power ratings on appliances work exactly this way — they tell you how much energy the device uses per second while running.
Energy joules (J)
÷ time →
Power watts (W)
1 W =
1 J per second
WE 3
A 1500 W motor does 24 000 J of useful work. How long does it take? (Assume it works at full power throughout.)
Step 1 — start from P = ΔW ÷ Δt and rearrange for time
Δt = ΔW ÷ P
Step 2 — substituteΔt = 24 000 ÷ 1500Δt = 16 sA more powerful motor would do the same 24 000 J of work in less time.
🛠️ Choosing the right power equation
Given work (or energy) and time? Use P = ΔW / Δt.
Given a force and a steady velocity? Use P = Fv.
Work not given directly? Find it first — often the GPE gained (mgΔh) or the KE gained.
Steady speed against resistance? The driving force equals the resistive force, so P = (resistive force) × v.
Check the direction — P = Fv needs force and velocity aligned.
💡 Top tips
Two equations, one idea.P = ΔW / Δt and P = Fv are both on the data sheet — pick the one that matches your data.
Drag force = driving force at steady speed. The force needed to calculate power equals the resistance you’re overcoming.
Watch the units. Convert km to m and minutes to seconds before substituting.
Same direction. If the force is at an angle to the velocity, use only the component along the motion.
Quick recap: Power is the rate of transferring energy, P = ΔW / Δt, measured in watts (1 W = 1 J s−1). For a constant force moving at steady velocity, P = Fv, with force and velocity in the same direction. The same work done in less time means more power.
⚠ Common mistakes
Confusing power with energy — power is energy per second, not the total energy
Forgetting to convert units (km to m, minutes to seconds) before calculating
Using P = Fv when the force and velocity are not aligned
Not realising the driving force equals the resistive force at constant speed
Forgetting to find the work done first when only mass, height or speed is given
Power leads straight into the idea of how well a machine uses the energy it’s given. No real device turns all its input into useful output — some is always wasted. The measure of how good it is at this is efficiency, which compares useful power (or energy) out to total power in. That’s exactly where we head next.
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