IB Physics HL A.3 Work, Energy and Power Question Bank
Practise IB Physics HL A.3 by solving work-energy, power, efficiency and energy-density problems across mechanical systems.
- Syllabus
- First assessment 2025
- Course
- Physics HL
- Level
- HL
Practise IB Physics HL A.3 by solving work-energy, power, efficiency and energy-density problems across mechanical systems.
Two sources of light produce an interference pattern on a screen.
Light of wavelength 720 nm is incident on two narrow slits that are separated by 0.12 mm . An interference pattern is observed on a screen. P1 and P2 are the points of destructive interference closest to the central maximum M .

Suggest how energy conservation is consistent with the fact that the energy at P1 and P2 is zero.
A beam of light containing all wavelengths in the range [550 nm,650 nm] is incident normally on a diffraction grating. The grating has 580 lines per mm . A diffraction pattern is observed on a screen as shown.
The energy missing at P1 and P2 is found at the maxima
This question is about the motion of a ship and observing objects from it.
Outline the meaning of work.
work done = force × distance moved;
(distance moved) in direction of force;
or
energy transferred; from one location to another;
or
work done =Fscosθ;
with each symbol defined;
Some cargo ships use kites working together with the ship's engines to move the vessel.

The tension in the cable that connects the kite to the ship is 250 kN . The kite is pulling the ship at an angle of 39∘ to the horizontal. The ship travels at a steady speed of 8.5 m s−1 when the ship's engines operate with a power output of 2.7 MW .
Calculate the work done on the ship by the kite when the ship travels a distance of 1.0 km .
horizontal force =250000×cos39∘(=1.94×105 N);
work done =1.9×108 J;
Show that, when the ship is travelling at a speed of 8.5 m s−1, the kite provides about 40 % of the total power required by the ship.
power provided by kite =(1.94×105×8.5=)1.7×106 W;
total power =(2.7+1.7)×106( W)(=4.4×106 W);
fraction provided by kite =2.7+1.71.7;
38 % or 0.38 ; (must see answer to 2+ sig figs as answer is given)
Marking guidance:
Allow answers in the range of 37 to 39 % due to early rounding.
or
Award [3 max] for a reverse argument such as:
if 2.7 MW is 60 %;
then kite power is 32×2.7MW=1.8MW;
shows that kite power is actually 1.7 MW ; (QED)
The kite is taken down and no longer produces a force on the ship. The resistive force F that opposes the motion of the ship is related to the speed v of the ship by
where k is a constant.
Show that, if the power output of the engines remains at 2.7 MW , the speed of the ship will decrease to about 7 ms−1. Assume that k is independent of whether the kite is in use or not.
P=(kv2)×v=kv3;
v2v1=(3(P2P1)=)3(4.42.7);
final speed of ship =7.2 ms−1; (at least 2 sig figs required).
Approximate answer given, marks are for working only.
The Sankey diagram shows the energy input from fuel that is eventually converted to useful domestic energy in the form of light in a filament lamp.

What is true for this Sankey diagram?
The overall efficiency of the process is 10 %.
Generation and transmission losses account for 55 % of the energy input.
Useful energy accounts for half of the transmission losses.
The energy loss in the power station equals the energy that leaves it.
A