Work, Energy, and Simple Machines — Detailed Notes
When forces are complicated, the ideas of work, energy and power give a simpler way to analyse motion. Simple machines let us do tasks with less effort. Energy is the capacity to do work.
7.1 Work done by a constant force
Work = force × displacement in the direction of the force, i.e. W = F × s. SI unit = joule (J); 1 J = 1 N × 1 m = 1 kg m² s⁻². On a force–displacement graph, work = the area under the graph.
Zero work: when F = 0, when s = 0 (e.g. pushing a rigid wall — you tire but do no work on it), or when the force is perpendicular to the displacement (e.g. carrying a box horizontally). Positive work — displacement in the direction of the force (pushing a wheelchair forward). Negative work — displacement opposite to the force (a goalkeeper stopping a ball).
7.2 Work–energy theorem
Work done on an object = change in its energy. Positive work increases energy; negative work decreases it. Energy can also be transferred as heat, radiation, sound, electrically, etc.
7.3 Forms of energy
Mechanical, thermal, light, sound, electrical, chemical, nuclear — and energy can be converted from one form to another (e.g. electrical → light in a bulb; chemical in food → mechanical in muscles).
7.4 Mechanical energy
Kinetic energy (energy of motion): K = ½mv² (derived from W = Fs with F = ma and v² = u² + 2as). Doubling the speed makes KE four times larger. Potential energy — energy stored by deformation (stretched band, spring, bow) or by the relative positions of interacting objects (gravitational, electric, magnetic). Gravitational PE near the Earth: U = mgh.
Conservation of mechanical energy: for an object moving only under gravity (no friction/air resistance), the sum KE + PE stays constant. As a body falls, PE converts to KE (total = mgh throughout). A pendulum and a slide show this; a child reaching the bottom of a slide of height h has v = √(2gh), independent of mass or slide shape.
7.5 Power
Power = rate of doing work: P = W / t. SI unit = watt (W); 1 W = 1 J s⁻¹. (1 horsepower = 746 W.) Same work done in less time needs more power.
7.6 Simple machines
Machines make work easier by changing the magnitude or direction of the force — but they do not reduce the total work (less force → larger distance). Mechanical advantage (MA) = load ÷ effort.
• Pulley — a fixed pulley only changes direction (MA = 1, convenient to pull down); a movable pulley / pulley system gives MA > 1. • Inclined plane — MA = length L ÷ height h (> 1); a longer, gentler ramp needs less effort. • Lever — a rigid bar turning about a fulcrum; effort × effort-arm = load × load-arm, so MA = effort-arm ÷ load-arm. Three classes: Class I fulcrum in middle (seesaw, scissors), Class II load in middle (wheelbarrow, bottle opener), Class III effort in middle (tongs, tweezers).
Key Terms
work (W = Fs), joule, positive/negative/zero work, work–energy theorem, kinetic energy (½mv²), potential energy (mgh), conservation of mechanical energy, power (P = W/t), watt, simple machine, effort, load, mechanical advantage, pulley, inclined plane, lever, fulcrum.
Revise, Reflect, Refine — All NCERT Exercise Answers
1State True or False: (i) work is done even if the object doesn’t move; (ii) lifting a bucket up is positive work; (iii) SI unit of work & energy is joule; (iv) a stretched motionless rubber band has kinetic energy; (v) energy can change form.
(i) False — if there is no displacement (s = 0), no work is done, however much force is applied. (ii) True — force and displacement are both upward, so the work is positive. (iii) True — both are measured in joule (J). (iv) False — a stretched but still rubber band has potential energy, not kinetic energy (it is not moving). (v) True — energy can be converted from one form to another.
2Fill in the blanks.
(i) Work done = force × displacement (in the direction of force). (ii) 1 joule is done when a force of 1 newton displaces an object by 1 metre in the direction of the force. (iii) Kinetic energy = ½mv². (iv) Potential energy at height h = mgh. (v) Power is the rate at which work is done.
3A ball thrown upward reaches its highest point. Which statements are correct?
(iii) Its kinetic energy is zero (momentary velocity = 0) and (iv) its potential energy is maximum (greatest height). The others are wrong: the force of gravity still acts on it, so the force is not zero (i) and the acceleration is g, not zero (ii).
4Identify the energy transformation in each case.
(i) Truck uphill — chemical (fuel) → kinetic + gravitational potential. (ii) Unwinding watch spring — elastic potential → kinetic. (iii) Photosynthesis — light → chemical. (iv) Water flowing from a dam — potential → kinetic. (v) Burning matchstick — chemical → heat + light. (vi) Firecracker explosion — chemical → heat + light + sound. (vii) Speaking into a microphone — sound → electrical. (viii) Glowing bulb — electrical → light + heat. (ix) Solar panel — light → electrical.
5Student m = 50 kg, building h = 72.5 m, g = 10. (i) PE gain lifted straight up (ii) PE gain climbing stairs (iii) conclusion about path.
(i) U = mgh = 50 × 10 × 72.5 = 36,250 J. (ii) Climbing the stairs to the same height gives the same U = 36,250 J. (iii) Gravitational potential energy depends only on the vertical height, not on the path taken — both routes give the same PE gain.
6A crane lifts a mass to the 10th floor in time t, then the same mass to the 20th floor in 2t. How much more energy and power are needed?
The 20th floor is twice the height of the 10th, so the work (energy) = mgh doubles → twice the energy. Power P = W/t: first case P₁ = W/t; second case P₂ = 2W/(2t) = W/t = same power. So twice the energy is needed, but the power required is unchanged.
7What determines the energy to raise a flag with a pulley? Does raising it slowly/quickly change the work? If the speed doubles, how does the power change?
The work (energy) depends on the flag’s weight (mg) and the height h of the pole — W = mgh (a fixed pulley only changes the direction of the effort, MA = 1). Raising it slowly or quickly makes no difference to the work done (same mgh). But power P = W/t: doubling the speed halves the time, so the power required doubles.
8A 60 kg man rides a 100 kg scooter to speed v. Next day his 40 kg son joins; same speed in the same time. Ratio of fuel used on the two days?
Fuel ≈ energy given = kinetic energy ½mv². Day 1 mass = 60 + 100 = 160 kg; Day 2 mass = 160 + 40 = 200 kg. For the same v, energy ∝ mass, so fuel₁ : fuel₂ = 160 : 200 = 4 : 5 (day 1 uses 4/5 of day 2’s fuel).
9A seesaw is balanced with a child on one side and an adult (twice the child’s weight) on the other. Draw the distances from the fulcrum.
For balance: effort × effort-arm = load × load-arm, i.e. weight × distance must be equal on both sides. Since the adult’s weight is twice the child’s, the adult must sit at half the distance from the fulcrum that the child sits. (Figure: child at distance 2d on one side, adult at distance d on the other side of the fulcrum — because W × 2d = 2W × d.)
10A 2 kg ball is thrown up at 20 m s⁻¹. (i) sign of work by gravity on the way up and down (ii) if it reaches 19.4 m, work done by air resistance (g = 10).
(i) Going up, gravity acts opposite to the motion → negative work; coming down, gravity acts along the motion → positive work. (ii) Initial KE = ½ × 2 × 20² = 400 J. Without air resistance max height = v²/2g = 20 m, but it only reaches 19.4 m. Work by gravity going up = −mgh = −(2 × 10 × 19.4) = −388 J. At the top KE = 0, so total work = −400 J. Work by air resistance = −400 − (−388) = −12 J (12 J lost to air resistance).
11A 10 kg block (negligible friction) has KE = 180 J at 0 m; a force acts over 0–4 m (Fig. 7.37). Find the speed (i) at 0 m and (ii) at 4 m. Any negative acceleration?
(i) ½mv² = 180 → v² = 360/10 = 36 → v = 6 m s⁻¹ at 0 m. Work done by the force = area under the F–s graph = 50 N × 4 m = 200 J. (ii) KE at 4 m = 180 + 200 = 380 J → v² = 760/10 = 76 → v ≈ 8.7 m s⁻¹. Since the force acts in the direction of motion (positive work) throughout, the block only speeds up — no negative acceleration.
12Moon’s gravity ≈ 1/6 of Earth’s. On Earth an astronaut throws a ball up to 8 m. With the same velocity, how high will it go on the Moon?
At the top v = 0, so (from v² = u² − 2gh) the height h = u²/(2g) — for the same throw velocity u, h is inversely proportional to g. The Moon’s g is 1/6 of Earth’s, so the height becomes 6 times: 8 × 6 = 48 m.
13A 1000 kg car: speed–time graph (Fig. 7.38), points A, B, C. (i) motion A→B (ii) KE at A (iii) work by brakes B→C (iv) what the KE becomes.
(i) From A to B the speed–time line is flat, so the car moves at constant speed (no acceleration). (ii) KE at A = ½mv², with v read from the flat part of the graph (e.g. if v = 30 m s⁻¹, KE = ½ × 1000 × 30² = 450,000 J). (iii) From B to C the car decelerates to rest, so the brakes do negative work equal to the car’s KE at B (e.g. −450,000 J). (iv) The kinetic energy is transformed mainly into heat (in the brakes, tyres and road) and some sound. (Substitute the speed read from the graph.)
14PE–displacement graph of a 0.5 kg ball on a frictionless track (Fig. 7.39). At O, v = 0 and PE = 30 J. Find the velocity at P, Q, R.
Total mechanical energy is conserved = 30 J. At any point KE = 30 − (PE there), and v = √(2·KE/m) = √(2·KE/0.5) = √(4·KE). Read the PE at each point from the graph: e.g. if PE = 20 J at R → KE = 10 J → v = √40 ≈ 6.3 m s⁻¹; if PE = 10 J at Q → KE = 20 J → v = √80 ≈ 8.9 m s⁻¹; if PE = 0 at P → KE = 30 J → v = √120 ≈ 11 m s⁻¹. (Lower PE → higher speed.)
15A 1.5 kg coconut falls from a 10 m tree onto wet sand (g = 10). (i) velocity just before hitting (ii) if sand resists with 3000 N, depth of the depression.
(i) v² = 2gh = 2 × 10 × 10 = 200 → v = √200 ≈ 14.1 m s⁻¹. (ii) Energy at impact = mgh = 1.5 × 10 × 10 = 150 J (= its KE). The sand’s resistive force does this much work over depth d: F × d = 150 → 3000 × d = 150 → d = 0.05 m = 5 cm.