Squares, Ratios and Rearranging Formulas
The kinetic energy formula Ek = ½ m v² has a square in it, and that square is behind some of the chapter's most surprising answers: double the speed and the energy becomes four times. You will also have to turn formulas around – find the height from the energy, or the time from the power – and read the work done as the area under a graph. This page practises all of that with numbers before the physics arrives.
- Multiplication tables and simple division.
- Units in kg, m, s, J and W (page 2).
1. Squaring a number
The square of a number is the number multiplied by itself: v² = v × v. So 5² = 25 and 12² = 144. Picture it as a square of tiles with v tiles along each side. Going backwards, the square root √ asks “which number times itself gives this?”: √144 = 12 and √196 = 14.
Watch the tile count: going from v = 5 to v = 10 doubles each side but makes four times as many tiles (25 → 100). This is why doubling a speed makes the kinetic energy four times bigger.
| v | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 20 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| v² | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 | 121 | 144 | 169 | 196 | 225 | 400 |
0.5² = 0.5 × 0.5 = 0.25 (smaller than 0.5!), and 2.5² = 6.25. Squaring a number less than 1 makes it smaller.
2. Working out ½ × m × v² in the right order
In ½ m v², only v is squared. Do it in this order:
- Square the speed: v².
- Multiply by the mass: m × v².
- Halve it (multiply by ½, or divide by 2).
The three numbers can be multiplied in any order (½ × m × v² = m × v² ÷ 2 = ½ × v² × m), but the square belongs to v alone. Squaring the mass or the ½ as well is the most common mistake.
Worked example 1 · A rolling stone
Q. A stone of mass 4 kg rolls down a hillside at 5 m/s. Find ½ × m × v² for it (its kinetic energy, in joules).
Square 5 first (25), then multiply by 4, then halve.
3. What happens when one quantity changes?
Many questions ask “what happens to the energy if the mass is doubled?” or “if the speed is tripled?”. You do not need to calculate everything again – just follow the factor:
- A quantity that appears plainly (like m in ½ m v², or h in m g h): the result changes by the same factor. Double m → double the energy.
- A quantity that is squared (like v in ½ m v²): the result changes by the square of the factor. Double v → 2² = 4 times; triple v → 3² = 9 times; halve v → (½)² = ¼.
Start: m = 10 kg, v = 2 m/s, so ½ × 10 × 2² = 20 J (the first row of the textbook's Table 6.4). The grey bar is the start; the coloured bar is the new value.
Plain quantity: factor k → result × k. Squared quantity: factor k → result × k². So in ½ m v²: double m → ×2, double v → ×4, triple v → ×9, halve v → ×¼.
4. Turning a formula around
A formula is a balance: whatever you do to one side you must do to the other. To get one letter on its own, undo what is being done to it – undo a multiplication by dividing, undo a division by multiplying.
h = E ÷ (m g). More energy → higher. A heavier object needs more energy for the same height, so the same energy lifts it less.
Worked example 2 · Finding a height
Q. A 2 kg flower pot on a shelf has 300 J of potential energy, E = m g h. How high is the shelf? (g = 10 m/s²)
Divide both sides of E = m g h by (m × g): h = E ÷ (m g). Work out m × g first.
5. Straight lines and curves on a graph
A graph (ഗ്രാഫ്) shows how one quantity changes with another. If y is simply a number times x (y = k x), the graph is a straight line through the origin: Ep = m g h against h is like this. If y depends on x² (y = k x²), the graph is a curve that gets steeper and steeper: Ek = ½ m v² against v is like this. Being able to tell them apart is exactly what “Let us assess” Q8 tests.
6. Work as the area under a graph
If a constant force F pushes an object through a displacement s, the force–displacement graph is a flat line at height F. The rectangle under it has width s and height F, so its area = F × s – which is exactly the work done. (“Let us assess” Q15 uses this.)
Each small square is 1 N tall and 0.1 m wide, so it stands for 1 × 0.1 = 0.1 J. Count the shaded squares and multiply by 0.1 J – you get F × s.
Worked example 3 · Area under a force–displacement graph
Q. A force–displacement graph is a horizontal line at 5 N from s = 0 to s = 0.6 m. Find the work done.
Work = area of the rectangle = height × width = F × s.