PhysicsICSE Class 10 · Machines

Mechanical Advantage, Velocity Ratio and Efficiency: η = MA/VR with Solved Numericals (ICSE Class 10)

Q: Why is the efficiency of a practical machine always less than 100%?

Some of the work put in is always used to overcome friction between moving parts and to lift parts of the machine itself, such as a pulley block. That energy is not useful output, so the work output is less than the work input.

Q: Does a bigger load change the velocity ratio of a machine?

No. VR depends only on the design (distances or the number of supporting strands). A bigger load can raise the MA and efficiency, because the fixed losses become a smaller share of the work.

Q: Can a machine with MA less than 1 be useful?

Yes. It needs more effort than the load, but the load moves farther and faster than the effort. Fire tongs and the human forearm are used this way, for reach, speed or convenience.

Updated 15 October 2026 · Fact-checked

Mechanical advantage is load divided by effort, MA = L/E. Velocity ratio is the distance moved by the effort divided by the distance moved by the load, VR = dE/dL. Efficiency is useful work output divided by work input, and η = MA/VR. In every practical machine η < 1, so MA < VR.

The effort end moves farther (dE) than the load end (dL), so VR is larger than MA, because η = MA / VR is always less than 1.

Illustration drawn for QuestionBuddy.

Understand Mechanical Advantage, Velocity Ratio and Efficiency

A machine lets a small force do a job that needs a larger force, or lets a force act at a more convenient place or in a more convenient direction. Three numbers describe how well it does this.

1. Mechanical advantage (MA). The ratio of the load overcome to the effort applied:

$M A = \frac{L}{E} .$

It is a ratio of two forces, so it has no unit. $M A > 1$ means the machine is a force multiplier; $M A < 1$ means it is used to gain speed or distance (the load moves farther than the effort); $M A = 1$ means it only changes the direction of the force.

2. Velocity ratio (VR). The ratio of the distance moved by the effort to the distance moved by the load in the same time:

$V R = \frac{d _{E}}{d _{L}} = \frac{V _{E}}{V _{L}} .$

Since both distances take the same time, the ratio of distances equals the ratio of speeds. VR also has no unit. It depends only on the design (the arm lengths of a lever, the number of strands in a pulley system), not on friction or on the load.

3. Efficiency (η). The ratio of the useful work done by the machine (work output) to the work done on it (work input):

$η = \frac{W _{o}}{W _{i}} = \frac{L \times d _{L}}{E \times d _{E}} .$

It has no unit and is often written as a percentage.

Deriving the relation between η, MA and VR (the syllabus asks for this derivation).

$η = \frac{L \times d _{L}}{E \times d _{E}} = \frac{L}{E} \times \frac{d _{L}}{d _{E}} = M A \times \frac{1}{V R} .$

So $η = \frac{M A}{V R}$ , or $M A = η \times V R$ .

Ideal and practical machines. In an ideal machine there is no friction and the moving parts are weightless, so all the input work becomes useful output: $η = 1$ and $M A = V R$ . In a practical machine part of the input work is always wasted, in overcoming friction and in lifting moving parts such as a pulley block or a lever's own weight. Hence $W_{o} < W_{i}$ , so $η < 1$ and $M A < V R$ . The wasted energy appears mostly as heat (and some sound).

Energy bookkeeping. Work input = useful work output + work wasted:

$W_{i} = W_{o} + W_{wasted} .$

No machine creates energy. A machine that gains force must lose distance: if $M A = 4$ (ideally), the effort must move four times as far as the load.

MA from a graph. If load is plotted on the y-axis against effort on the x-axis, the graph is a straight line whose slope is $\frac{Δ L}{Δ E} = M A$ . If the axes are swapped (effort on the y-axis), the slope is $\frac{1}{M A}$ .

What changes η? Efficiency falls with more friction and with heavier moving parts. For a given machine it rises as the load increases, because the fixed waste (such as lifting the pulley block) becomes a smaller fraction of the total work.

Mechanical Advantage, Velocity Ratio and Efficiency: formulas and key facts

Mechanical advantage: $M A = \frac{L}{E}$; No unit. Write as a decimal, not a fraction.
Velocity ratio: $V R = \frac{d _{E}}{d _{L}} = \frac{V _{E}}{V _{L}}$; No unit; fixed by the design of the machine.
Efficiency: $η = \frac{W _{o}}{W _{i}} = \frac{L \times d _{L}}{E \times d _{E}}$; Multiply by 100 for a percentage.
Relation between η, MA and VR: $η = \frac{M A}{V R} \Rightarrow M A = η \times V R$; Use η as a decimal (80% = 0.8) in this formula.
Energy balance: $W_{i} = W_{o} + W_{wasted}$; Practical machine: η < 1 and MA < VR.

How to solve Mechanical Advantage, Velocity Ratio and Efficiency questions

Almost every machine numerical is solved by writing down the three ratios and the link η = MA/VR, then filling in what is given.

1List the data: L, E, $d_{L}$ , $d_{E}$ , VR or η. Keep forces in one unit (all N or all kgf).
2Find whichever ratios come directly: $M A = \frac{L}{E}$ , $V R = \frac{d _{E}}{d _{L}}$ .
3Convert a percentage efficiency to a decimal before using $M A = η \times V R$ .
4For the effort in a practical machine, use $E = \frac{L}{M A}$ with the practical MA, never $\frac{L}{V R}$ .
5For work, use $W_{i} = E \times d_{E}$ and $W_{o} = L \times d_{L}$ ; the wasted energy is $W_{i} - W_{o}$ .
6Check: η must be less than 1 (or 100%) and MA less than VR for a real machine.

Mechanical Advantage, Velocity Ratio and Efficiency in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A pulley system on a building site

Suppose a worker uses a pulley system of velocity ratio 4 and efficiency 80% to raise a 320 N tub of mortar through 1.5 m to a scaffold.

Find the effort she must apply and the length of rope she pulls through.

Try it first, then show the solutionSolution

$M A = η \times V R = 0.8 \times 4 = 3.2$ .
$E = \frac{L}{M A} = \frac{320}{3.2} = 100$ N.
Rope pulled $= V R \times d_{L} = 4 \times 1.5 = 6$ m.

Answer: 100 N; 6 m of rope.

Memory hook: Friction changes the effort, never the rope length: VR fixes the distances.

A machine with MA less than 1

Imagine a pair of sugar tongs that acts as an ideal lever with MA 0.4, holding a 2 N sweet.

Find the effort needed, and how far the tips move when the fingers move 1 cm. Why use such a machine?

Try it first, then show the solutionSolution

$E = \frac{L}{M A} = \frac{2}{0.4} = 5$ N.
Ideal lever: $V R = M A = 0.4$ , so $d_{L} = \frac{d _{E}}{V R} = \frac{1}{0.4} = 2.5$ cm.

Answer: 5 N; the tips move 2.5 cm. The tongs trade force for distance and reach, and keep fingers off the food.

Memory hook: MA below 1 is not a bad machine: it buys distance, speed or convenience.

Tricks to remember Mechanical Advantage, Velocity Ratio and Efficiency

Short memory aids. Say each one aloud once, then use it on a question the same day.

MA has Load on top, VR has Effort on top
MA puts the Load on top; VR puts the Effort on top. They are upside-down twins.
Use it when: Writing the definitions without mixing them up.
η = MA ÷ VR
'Efficiency is MA per VR': what you actually get divided by what the design promises.
Use it when: The relation, and remembering MA < VR when η < 1.
No units, no fractions
MA, VR and η are pure numbers: write 1.6, not 8/5, and never 'N'.
Use it when: Avoiding the errors examiners reported in 2024.
Slope reads y over x
Load up the y-axis, effort along the x-axis: rise over run is load over effort, which is MA.
Use it when: Load–effort graph questions.

Mechanical Advantage, Velocity Ratio and Efficiency: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Efficiency from MA and VR · 1 mark · MCQBoard level

A machine has a mechanical advantage of 3 and a velocity ratio of 4. Its efficiency is:
(a) 75%
(b) 133%
(c) 12%
(d) 25%

Try it first, then show the solutionSolution

$η = \frac{M A}{V R} = \frac{3}{4} = 0.75 = 75%$ .

Answer: (a) 75%

Example 2: All three ratios from measurements · 3 marks · Short answerBoard level

An effort of 250 N moves through 2 m and raises a load of 800 N through 0.5 m. Find the mechanical advantage, the velocity ratio and the efficiency of the machine.

Try it first, then show the solutionSolution

$M A = \frac{L}{E} = \frac{800}{250} = 3.2$ . (1)
$V R = \frac{d _{E}}{d _{L}} = \frac{2}{0.5} = 4$ . (1)
$η = \frac{M A}{V R} = \frac{3.2}{4} = 0.8 = 80%$ . Check: $\frac{W _{o}}{W _{i}} = \frac{800 \times 0.5}{250 \times 2} = \frac{400}{500} = 0.8$ . (1)

Answer: MA = 3.2, VR = 4, η = 80%

Example 3: Effort in kgf from efficiency · 3 marks · Short answerHarder board level

A machine of velocity ratio 5 works at 70% efficiency. It is used to lift a load of 140 kgf.
(a) Find its mechanical advantage.
(b) Find the effort needed, in kgf and in newtons. (1 kgf = 9.8 N)

Try it first, then show the solutionSolution

(a) $M A = η \times V R = 0.7 \times 5 = 3.5$ . (1)
(b) $E = \frac{L}{M A} = \frac{140}{3.5} = 40$ kgf. (1)
In newtons: $40 \times 9.8 = 392$ N. (1)

Answer: (a) 3.5 (b) 40 kgf = 392 N

Example 4: Reading a load–effort graph · 2 marks · Short answerHarder board level

For a machine of velocity ratio 4, a graph of load (y-axis) against effort (x-axis) is a straight line through the origin and the point (12 N, 42 N). Find (a) the mechanical advantage, (b) the efficiency.

Try it first, then show the solutionSolution

(a) Slope $= \frac{Δ L}{Δ E} = \frac{42 - 0}{12 - 0} = 3.5 = M A$ . (1)
(b) $η = \frac{3.5}{4} = 0.875 = 87.5%$ . (1)

Answer: (a) 3.5 (b) 87.5%

Example 5: Full energy account · 4 marks · Long answerHarder board level

A machine of velocity ratio 8 and efficiency 60% raises a load of 1200 N through 1.5 m. Find (a) its MA, (b) the effort, (c) the distance moved by the effort, (d) the work input, (e) the energy wasted.

Try it first, then show the solutionSolution

(a) $M A = 0.6 \times 8 = 4.8$ . (½)
(b) $E = \frac{1200}{4.8} = 250$ N. (1)
(c) $d_{E} = V R \times d_{L} = 8 \times 1.5 = 12$ m. (½)
(d) $W_{i} = E \times d_{E} = 250 \times 12 = 3000$ J. (1)
(e) $W_{o} = 1200 \times 1.5 = 1800$ J, so wasted $= 3000 - 1800 = 1200$ J. (1)

Answer: (a) 4.8 (b) 250 N (c) 12 m (d) 3000 J (e) 1200 J

Example 6: Case study: a car jack · 4 marks · Case-basedHOTS / competency

In one stroke of a car jack, a driver pushes the handle with an effort of 50 N through 1.2 m, and the side of the car (a load of 6000 N) rises by 4 mm.
(a) Find the VR and the MA of the jack.
(b) Find its efficiency.
(c) How much energy is wasted per stroke?
(d) The efficiency is low. Why is this acceptable, and even useful, for a jack?

Try it first, then show the solutionSolution

(a) $d_{L} = 4 mm = 0.004$ m. $V R = \frac{1.2}{0.004} = 300$ ; $M A = \frac{6000}{50} = 120$ . (1)
(b) $η = \frac{120}{300} = 0.4 = 40%$ . (1)
(c) $W_{i} = 50 \times 1.2 = 60$ J; $W_{o} = 6000 \times 0.004 = 24$ J; wasted $= 36$ J. (1)
(d) The jack's job is to multiply force (MA = 120), and the energy lost per stroke is small. The large friction also stops the car sinking back when the driver lets go of the handle. (1)

Answer: (a) VR = 300, MA = 120 (b) 40% (c) 36 J (d) Force gain matters more than energy; friction holds the load in place.

Where students lose marks in Mechanical Advantage, Velocity Ratio and Efficiency

The slips examiners see most often, and what to write instead.

Mistake: Leaving MA as a fraction or giving it a unit
MA is a pure number and the scheme expects a decimal.
Wrong working
$M A = \frac{8}{5}$ N.
Right working
$M A = 1.6$ (no unit).
Do this: Divide out every ratio and write no unit after MA, VR or η.
Usually costs: Reported by CISCE examiners for 2024 Q6(iii).
Mistake: Using the ideal MA when the efficiency is given
A practical machine's MA is η × VR, which is less than VR.
Wrong working
VR 5, η 80%, load 400 N: $E = \frac{400}{5} = 80$ N.
Right working
$M A = 0.8 \times 5 = 4$ , so $E = \frac{400}{4} = 100$ N.
Do this: Find the practical MA first, then $E = \frac{L}{M A}$ .
Usually costs: Also reported for 2024 Q6(iii); usually loses the method and answer marks.
Mistake: Reading the load–effort graph upside down
Slope is change in y divided by change in x.
Wrong working
Load on the y-axis, effort on the x-axis, so slope $= \frac{1}{M A}$ .
Right working
Slope $= \frac{Δ L}{Δ E} = M A$ .
Do this: Name the axes before you name the slope.
Usually costs: The MCQ mark in 2025 Q1(vi), where candidates chose 1/MA.
Mistake: Mixing kgf and N, or percentages and decimals
MA = L/E needs both forces in the same unit, and η × VR needs η as a decimal.
Wrong working
$M A = 80 \times 2 = 160$ .
Right working
$M A = 0.8 \times 2 = 1.6$ .
Do this: Convert units on a separate line before substituting.
Usually costs: The answer mark; kgf/N confusion was noted in 2024.

Exam tips

In 2024 Q6(iii) (VR 2, load 48 kgf, η 80%) the expected working was $\frac{80}{100} = \frac{M A}{2}$ , so $M A = 1.6$ , then $1.6 = \frac{48}{E}$ , so $E = 30$ kgf. Show both formulas before substituting.
Learn the derivation of $η = \frac{M A}{V R}$ in three lines: define η, write $W_{o} = L \times d_{L}$ and $W_{i} = E \times d_{E}$ , then split into $\frac{L}{E} \times \frac{d _{L}}{d _{E}}$ .
When asked why MA < VR, give the reason in energy terms: some input work is wasted against friction and in lifting the moving parts, so η < 1.
Keep the answer in the unit of the question: if the load is in kgf, the effort can be left in kgf.

Practice questions on Mechanical Advantage, Velocity Ratio and Efficiency

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The SI unit of mechanical advantage is:
(a) newton
(b) joule
(c) N m⁻¹
(d) it has no unit
Hint
MA is a ratio of two forces.
Answer
(d) it has no unit
Q2 · 3 marks · Short answerBoard level
With an effort of 125 N moving through 1.8 m, a machine raises a load of 600 N through 0.3 m. Calculate its MA, VR and efficiency.
Hint
Find MA and VR first, then divide.
Answer
$M A = \frac{600}{125} = 4.8$ ; $V R = \frac{1.8}{0.3} = 6$ ; $η = \frac{4.8}{6} = 0.8 = 80%$ .
Q3 · 2 marks · Short answerHarder board level
A machine has an efficiency of 75% and a mechanical advantage of 6. (a) Find its velocity ratio. (b) How far must the effort move to raise the load by 0.2 m?
Hint
$V R = \frac{M A}{η}$ .
Answer
(a) $V R = \frac{6}{0.75} = 8$ . (b) $d_{E} = 8 \times 0.2 = 1.6$ m.
Q4 · 2 marks · Short answerHOTS / competency
A student reports that her model machine has MA = 5 and VR = 4. Without repeating the experiment, explain why the result must be wrong.
Hint
Work out the efficiency her figures imply.
Answer
Her figures give $η = \frac{5}{4} = 1.25$ , i.e. 125%: the machine would give out more work than was put in, which breaks conservation of energy. For any real machine $η < 1$ , so MA must be less than VR.
Q5 · 2 marks · Short answerHarder board level
Machine A has VR 4 and MA 3. Machine B has VR 6 and MA 4.2. (a) Which needs less effort for the same load? (b) Which is more efficient?
Hint
Effort depends on MA; efficiency on MA ÷ VR.
Answer
(a) B, since its MA is larger. (b) $η_{A} = \frac{3}{4} = 75%$ , $η_{B} = \frac{4.2}{6} = 70%$ : A is more efficient.
Q6 · 3 marks · Short answerHarder board level
2500 J of work is done on a machine of efficiency 84%. It lifts a load of 700 N. Find (a) the useful work output, (b) the energy wasted, (c) the height through which the load is raised.
Hint
$W_{o} = η \times W_{i}$ , then $h = \frac{W _{o}}{L}$ .
Answer
(a) $W_{o} = 0.84 \times 2500 = 2100$ J. (b) $2500 - 2100 = 400$ J. (c) $h = \frac{2100}{700} = 3$ m.
Q7 · 4 marks · Long answerHOTS / competency
A machine has VR 5. Without friction, a load of 450 N could be raised with an effort of 90 N, but friction means 100 N is actually needed. (a) Find the actual MA and the efficiency. (b) What part of the effort is used only to overcome friction? (c) If the friction were reduced, which of MA, VR and η would change?
Hint
The ideal effort is $\frac{L}{V R}$ .
Answer
(a) $M A = \frac{450}{100} = 4.5$ ; $η = \frac{4.5}{5} = 0.9 = 90%$ . (b) $100 - 90 = 10$ N. (c) MA and η would increase; VR would not change, because it depends only on the design.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.MA of a machine that lifts 600 N with an effort of 200 N?Show answer
3.
2.MA of a machine with VR 3 and efficiency 90%?Show answer
2.7.
3.Which of MA, VR and η depends only on the design of the machine?Show answer
VR.
4.Efficiency of an ideal machine?Show answer
1 (100%), with MA = VR.
5.Slope of a graph of load (y) against effort (x)?Show answer
The mechanical advantage.

Mechanical Advantage, Velocity Ratio and Efficiency: frequently asked questions

Why is the efficiency of a practical machine always less than 100%?

Some of the work put in is always used to overcome friction between moving parts and to lift parts of the machine itself, such as a pulley block. That energy is not useful output, so the work output is less than the work input.

Does a bigger load change the velocity ratio of a machine?

No. VR depends only on the design (distances or the number of supporting strands). A bigger load can raise the MA and efficiency, because the fixed losses become a smaller share of the work.

Can a machine with MA less than 1 be useful?

Yes. It needs more effort than the load, but the load moves farther and faster than the effort. Fire tongs and the human forearm are used this way, for reach, speed or convenience.