Class 9 Science: Work, Energy, and Simple Machines — Chapter Question Set

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Work done by a constant force is the product of the magnitude of the force and the displacement of the object in the direction of the force. Its SI unit is the joule (J), where 1 J is the work done when a force of 1 N displaces an object by 1 m, so 1 J = 1 N m.

Work done is zero when the applied force is zero, or when the displacement is zero, or when the force is perpendicular to the displacement. Two examples are pushing a heavy wall without moving it (zero displacement) and carrying a box horizontally while applying an upward force (force perpendicular to displacement).

Work is positive when the force and displacement are in the same direction, as when a person pushes a wheelchair forward. Work is negative when the force and displacement are in opposite directions, as when a goalkeeper's hand applies a stopping force on an approaching ball.

The work-energy theorem states that the work done on an object or a system is equal to the change in its energy. When positive work is done the energy increases, and when negative work is done the energy decreases.

Kinetic energy is the energy possessed by an object due to its motion. It depends on the mass and velocity of the object and is given by the formula K = ½mv², where m is the mass and v is the velocity of the object.

Kinetic energy is proportional to the square of the velocity (K ∝ v²). Therefore, when the velocity is doubled, the kinetic energy becomes 2² = 4 times its original value.

Potential energy is the energy stored in an object due to its position, shape or the deformation of a system. For an object at a height h above the ground, the gravitational potential energy is given by U = mgh, where m is the mass and g is the acceleration due to gravity.

Power is the rate at which work is done, or the amount of work done per unit time: P = W/t. Its SI unit is the watt (W), where 1 W = 1 J/s.

The mechanical advantage of a simple machine is the ratio of the load to the effort: MA = load/effort. It tells us how many times the machine multiplies the applied effort force.

A simple machine is a device that makes work easier by changing the magnitude or the direction of the applied force. The three simple machines studied in this chapter are the pulley, the inclined plane and the lever.

A fixed pulley only changes the direction of the applied effort, not its magnitude. Since the effort equals the load in this arrangement, the mechanical advantage MA = load/effort = 1.

If no external force other than gravity acts on an object, its total mechanical energy — the sum of kinetic energy and potential energy — remains constant throughout its motion. Energy may change form between kinetic and potential, but the total does not change.

Here m = 200 g = 0.2 kg, h = 10 m and g = 10 m s⁻². Potential energy PE = mgh = 0.2 × 10 × 10 = 20 J.

Work done W = mgh = 75 × 10 × 2 = 1500 J. Power P = W/t = 1500/5 = 300 W.

Since their kinetic energies are equal, ½mv₁² = ½(4m)v₂², which gives v₁² = 4v₂². Hence v₁ : v₂ = 2 : 1.

Here the force opposes the displacement, so W = F × s = 200 N × (−0.15 m) = −30 J. The negative sign indicates that the work done by the goalkeeper's hand is negative.

Hill roads are made to wind around so that the inclined plane becomes longer and the slope gentler. Since a longer inclined plane needs a smaller effort to climb the same height, the same amount of work is spread over a greater distance, making climbing easier.

A lever balances when the moment of the effort equals the moment of the load about the fulcrum, that is, effort × effort arm = load × load arm.

The kinetic energy gained = ½ × 1000 × 20² = 200000 J. This equals the work done by the engine, so the power = 200000/10 = 20000 W (or 20 kW).

Using conservation of energy, ½mv² = mgh, so v = √(2gh) = √(2 × 10 × 10) = √200 ≈ 14.14 m s⁻¹.

Introduction: Work depends on both the applied force and the displacement it produces in its direction.
Key Points:
1. Definition — Work = force × displacement in the direction of the force (W = F × s).
2. Positive work — When displacement is along the force (e.g. pushing a wheelchair forward).
3. Negative / zero work — When displacement is opposite to force (stopping a ball) or when s = 0 or force ⊥ displacement.
Conclusion: Thus the sign of work depends on the angle between the force and the displacement directions.

Introduction: The work-energy theorem and kinematic equations together give a formula for kinetic energy.
Key Points:
1. Kinematic link — From v² = u² + 2as, the displacement s = (v² − u²)/(2a).
2. Work done — W = F × s = ma × (v² − u²)/(2a) = ½m(v² − u²).
3. KE formula — Starting from rest (u = 0), the work done equals the KE gained: K = ½mv².
Conclusion: This derivation shows kinetic energy equals the work needed to accelerate a body from rest to velocity v.

Introduction: Potential energy is stored energy due to position or deformation within a system of interacting objects.
Key Points:
1. Definition — PE is the energy stored by an object due to its deformation or position in a system of objects.
2. Work to lift — To raise mass m slowly to height h, an upward force mg must be applied through distance h.
3. Derivation — By the work-energy theorem, W = F × s = mg × h = mgh, which is stored as gravitational PE.
Conclusion: So the gravitational potential energy of an object at height h is U = mgh, measured in joules.

Introduction: When only gravity acts, the total mechanical energy of a falling object remains constant at every point.
Key Points:
1. At the top — Object at rest at height h has PE = mgh and KE = 0; total ME = mgh.
2. During fall — PE decreases as height drops, but KE increases by exactly the same amount.
3. At the bottom — PE = 0 and KE = mgh; the total ME is still mgh, unchanged throughout.
Conclusion: This proves the mechanical energy is conserved when no external forces other than gravity act.

Introduction: Power measures how quickly work is done by a force on an object.
Key Points:
1. Definition — Power = work done / time taken; its SI unit is the watt (W = J/s).
2. Work done — W = mgh = 75 × 10 × 2 = 1500 J to lift the mass.
3. Power — P = W/t = 1500 J / 5 s = 300 W.
Conclusion: The weightlifter requires 300 W of power to lift the 75 kg mass in 5 s.

Introduction: An inclined plane reduces the effort needed to raise a load to a height by increasing the distance travelled.
Key Points:
1. Setup — A load of weight mg is pushed up a plane of length L to reach height h with effort F'.
2. Energy method — Ignoring friction, F' × L = mgh (work done = potential energy gained).
3. MA formula — MA = load/effort = mg/F' = L/h, which is greater than 1 since L > h.
Conclusion: A longer, gentler ramp gives a larger MA and reduces the effort required to lift the same load.

Introduction: A lever is a rigid bar that rotates about a fixed point called the fulcrum to lift a load with less effort.
Key Points:
1. Parts — A lever has a fulcrum, load (with load arm), and effort (with effort arm).
2. Balance condition — At balance, effort × effort arm = load × load arm (work is conserved).
3. MA formula — MA = load/effort = effort arm / load arm, so a longer effort arm gives more MA.
Conclusion: A lever reduces the effort but increases the distance moved, keeping the total work done the same.

Introduction: Energy conservation relates the PE at the top to the KE at the bottom of the fall.
Key Points:
1. PE at top — U = mgh = 0.2 × 10 × 10 = 20 J.
2. Energy conversion — All PE converts to KE at the bottom: ½mv² = mgh.
3. Velocity — v = √(2gh) = √(2 × 10 × 10) = √200 ≈ 14.14 m s⁻¹.
Conclusion: The ball has 20 J of potential energy at 10 m and strikes the ground at about 14.14 m s⁻¹.

Introduction: Energy conservation and the work done by the sand together determine the stopping distance on the ramp.
Key Points:
1. Initial KE — v = 20 m s⁻¹; KE = ½ × 10000 × 20² = 2000000 J.
2. Energy balance — Final KE = 0, final PE = mg(h) = 50000d; work by sand = −50000d; net: 2000000 = 100000d.
3. Ramp length — d = 2000000/100000 = 20 m.
Conclusion: The minimum length of the escape ramp needed to stop the truck is 20 m.

Introduction: A simple pendulum demonstrates the interconversion of potential and kinetic energy during oscillation.
Key Points:
1. At extremes — At points P and R, the bob momentarily stops: all energy is potential (mgh).
2. At the lowest point — At Q, the PE is zero and all energy is kinetic (½mv²).
3. Why it stops — Friction at the support and air resistance gradually convert ME into heat, stopping the bob.
Conclusion: In an ideal pendulum ME is conserved; in reality, energy losses cause the oscillations to die out.

Introduction: The engine's work equals the kinetic energy gained, and power is this work divided by the time taken.
Key Points:
1. KE gained — W = ½mv² = ½ × 1000 × 20² = 200000 J.
2. Time taken — The car reaches this speed in t = 10 s.
3. Power — P = W/t = 200000/10 = 20000 W.
Conclusion: The engine does 200000 J of work and delivers 20000 W (20 kW) of power during the acceleration.

Introduction: Energy conservation links the initial kinetic energy to the potential energy at the maximum height.
Key Points:
1. Initial KE — K = ½mv² = ½ × 2 × 20² = 400 J.
2. Max height — At the top, all KE → PE: mgh = 400 → h = 400/(2 × 10) = 20 m.
3. PE at top — U = mgh = 2 × 10 × 20 = 400 J, equal to the initial KE.
Conclusion: The ball starts with 400 J of KE, rises 20 m, and has 400 J of PE at the highest point.

Introduction: Winding hill roads are practical applications of the inclined plane studied in this chapter.
Key Points:
1. Inclined plane — A hill road is an inclined plane; MA = L/h, where L is the road length and h the height.
2. Gentle slope — Winding increases L, giving a larger MA so a smaller force (effort) is needed.
3. Trade-off — The distance travelled increases, but the total work done (mgh) remains the same.
Conclusion: Winding roads make hill climbs possible for vehicles by reducing the effort force required.

Introduction: The work done by the goalkeeper is negative because the force opposes the ball's displacement.
Key Points:
1. Values — F = 200 N opposite to the motion; displacement s = −0.15 m (opposite to force).
2. Calculation — W = F × s = 200 × (−0.15) = −30 J.
3. Sign explanation — The work is negative because the force and displacement are in opposite directions.
Conclusion: The goalkeeper does −30 J of work on the ball, removing 30 J of the ball's kinetic energy.

Introduction: Conservation of mechanical energy on a frictionless slide yields a speed independent of the child's mass.
Key Points:
1. At the top — The child starts from rest at height h, so PE = mgh and KE = 0.
2. At the bottom — All PE converts to KE: ½mv² = mgh.
3. Speed formula — Cancelling m: v = √(2gh), which depends only on h and g, not on mass.
Conclusion: So children of any mass reach the same speed at the bottom of a frictionless slide of the same height.

(a) Speed v = 154.8 km h⁻¹ = 43 m s⁻¹; KE = ½ × 0.2 × 43² = ½ × 0.2 × 1849 = 184.9 J.
(b) PE at the top = mgh = 0.2 × 10 × 10 = 20 J.
(c) KE is maximum and PE is minimum at the moment of release (the lowest point), because the ball is fastest there and at the lowest height.
(d) The total mechanical energy is conserved throughout the throw; as the ball rises, its kinetic energy is gradually converted into potential energy, keeping the sum constant.

(a) The 20th floor is at double the height, so the PE gained — and hence the energy required — is twice that for the 10th floor.
(b) Since power = energy/time and both the energy and the time double, the power required is the same in the two cases.
(c) With equal floor heights, the ratio of the heights (20th : 10th) is 2 : 1.
(d) No — potential energy depends only on the height gained, not on the path taken by the crane.

(a) Speed v = 72 km h⁻¹ = 20 m s⁻¹; KE = ½ × 10000 × 20² = 2000000 J.
(b) Work done by the sand equals the change in total energy: 2000000 − 50000d = 50000d, where 50000d is the PE gained as the truck rises along the ramp.
(c) Solving, 2000000 = 100000d, so d = 20 m.
(d) The truck's kinetic energy is converted partly into gravitational potential energy (the truck rises up the ramp) and partly into thermal energy (heat) due to friction with the sand.

(a) The 30 kg child should sit at seat D, 1 m from the fulcrum, since 15 × 2 = 30 × 1.
(b) The condition used is effort × effort arm = load × load arm (equal moments about the fulcrum).
(c) MA = load/effort = 30/15 = 2, which is also equal to the ratio effort arm/load arm = 2/1.
(d) No — the seesaw reduces the effort needed but increases the distance through which the effort moves, so the total work done remains the same.

(a) In the constant-speed phase the car moves with uniform speed, so its acceleration is zero and the net force on it is zero.
(b) KE = ½ × 1000 × 35² = 612500 J.
(c) The brakes do negative work equal to −612500 J, which removes all the kinetic energy of the car.
(d) The kinetic energy is transformed into thermal energy (heat) produced by friction between the brake pads, the discs/tyres and the road.

(a) Negative — gravity acts downward while the displacement during the upward motion is upward, so the two are opposite.
(b) Positive — during the downward motion both gravity and the displacement act downward, in the same direction.
(c) Expected PE at 20 m = 2 × 10 × 20 = 400 J; actual PE at 19.4 m = 2 × 10 × 19.4 = 388 J; so the work done by air resistance = 388 − 400 = −12 J.
(d) 12 J of mechanical energy was lost and converted into thermal energy (heat) due to air resistance.

(a) v = √(2gh) = √(2 × 10 × 10) = √200 ≈ 14.14 m s⁻¹.
(b) mgh = F_sand × d, i.e. 1.5 × 10 × 10 = 3000 × d, where h = 10 m and F_sand = 3000 N.
(c) d = (1.5 × 10 × 10)/3000 = 150/3000 = 0.05 m = 5 cm.
(d) The coconut's kinetic energy is completely converted into the work done against the sand (deformation of the sand) and into thermal energy (heat).

(a) Since the total mechanical energy (30 J) is conserved, the kinetic energy at any point is KE = 30 − PE.
(b) At P, KE = 30 − 20 = 10 J; so v = √(2 × KE/m) = √(2 × 10/0.5) = √40 ≈ 6.32 m s⁻¹.
(c) At Q, KE = 30 − 0 = 30 J; so v = √(2 × 30/0.5) = √120 ≈ 10.95 m s⁻¹.
(d) Kinetic energy cannot be negative, so the ball cannot enter a region where PE exceeds 30 J; at such a point the ball would stop and reverse direction (the turning point).

(a) Day 1: KE₁ = ½ × 160 × v² = 80v²; day 2: KE₂ = ½ × 200 × v² = 100v².
(b) The ratio KE₁ : KE₂ = 80 : 100 = 4 : 5.
(c) Assuming no losses, the fuel used is in the same ratio as the KE gained, so the ratio is 4 : 5.
(d) The work-energy theorem allows this — the chemical energy of the fuel is converted into the kinetic energy of the scooter.

(a) PE is maximum at A, the highest point of the track, and KE is maximum at the lowest point of the track.
(b) Each subsequent point is lower because some energy is lost to friction at every stage, so the ball cannot rise back to its original height.
(c) At any lower point the total mechanical energy ME = KE + PE = mgh, decreasing only because of the energy lost to friction.
(d) On a frictionless track the ball would reach exactly the same height at every peak and would oscillate forever, since its mechanical energy would be conserved.

Introduction: Work done by a force depends on the force applied and the displacement it produces in its direction.
Key Points:
1. Definition — Work = force × displacement in the direction of the force: W = F × s.
2. Positive work — Force and displacement are in the same direction (e.g. pushing a wheelchair forward).
3. Negative work — Force and displacement are in opposite directions (e.g. a goalkeeper stopping a ball).
4. Zero work — No displacement (pushing a wall), no force, or force perpendicular to displacement (carrying a box horizontally).
5. Unit — The SI unit is the joule (J); 1 J = 1 N × 1 m = 1 kg m² s⁻².
Conclusion: Thus, work depends on the angle between the force and the displacement, and can be positive, negative, or zero.

Introduction: The work-energy theorem connects the work done on an object to the change in its kinetic energy.
Key Points:
1. Theorem — The work done by a net force on an object equals the change in the object's energy.
2. Kinematic step — From v² = u² + 2as, the displacement is s = (v² − u²)/(2a).
3. Work substitution — W = F × s = ma × (v² − u²)/(2a) = ½m(v² − u²).
4. Starting from rest — If u = 0, the work done equals the final KE: K = ½mv².
5. Interpretation — Positive work increases KE; negative work decreases it; zero work keeps KE constant.
Conclusion: This derivation shows that kinetic energy is precisely the work required to accelerate a body from rest to speed v.

Introduction: Potential energy and its conservation are central ideas for understanding energy changes during motion.
Key Points:
1. Definition — PE is the energy stored by an object due to its deformation or position within a system.
2. Derivation of U = mgh — Work to lift mass m to height h at constant speed: W = F × s = mg × h = mgh.
3. Conservation principle — If only gravity acts, the total ME (KE + PE) stays constant at mgh throughout the fall.
4. At the top — PE = mgh, KE = 0; total ME = mgh.
5. At the bottom — PE = 0, KE = mgh; total ME is still mgh.
Conclusion: This proves that in free fall, potential energy converts into kinetic energy while total mechanical energy is conserved.

Introduction: Power combines the concept of work with the time taken to perform it, making it a key practical quantity.
Key Points:
1. Definition — Power = work done / time taken (P = W/t); SI unit: watt (1 W = 1 J/s).
2. Work done — v = 20 m s⁻¹; W = ½mv² = ½ × 1000 × 400 = 200000 J.
3. Power in 10 s — P = 200000/10 = 20000 W or 20 kW.
4. Power in 5 s — P = 200000/5 = 40000 W — twice as much power is needed.
5. Insight — Doing the same work in half the time requires double the power output.
Conclusion: This example shows that the same work done in different times requires very different power outputs.

Introduction: Simple machines make tasks easier by changing the magnitude or direction of effort without reducing total work.
Key Points:
1. Pulley — A wheel with a groove guiding a rope; a fixed pulley changes only direction, so MA = 1.
2. Inclined plane — A ramp: MA = L/h (load/effort); longer ramps give larger MA and smaller effort.
3. Lever — A rigid bar rotating about a fulcrum; MA = effort arm / load arm; longer effort arm gives more MA.
4. Common principle — All three reduce effort but increase the distance moved, keeping total work the same.
5. Formulae — Pulley: MA = 1; Inclined plane: MA = L/h; Lever: MA = (effort arm)/(load arm).
Conclusion: Simple machines trade distance for force, making physically demanding tasks manageable without reducing work.

Introduction: A pendulum is a classic demonstration of the interconversion of potential and kinetic energy.
Key Points:
1. At extreme P — The bob is at maximum height; all energy is PE = mgh, and KE = 0 momentarily.
2. At lowest point Q — PE = 0 and all energy is KE = ½mv²; the bob moves fastest here.
3. At extreme R — KE = 0 again and PE returns to mgh; the bob reaches the same height as at P.
4. Ideal case — In the absence of friction and air resistance, ME is perfectly conserved and oscillation continues forever.
5. Real case — Friction at the support and air resistance convert ME into heat, so the bob gradually loses height and stops.
Conclusion: Thus, energy conservation governs the pendulum, while real-world friction explains why it eventually comes to rest.

Introduction: The work-energy theorem relates the negative work done by the wire to the loss in the aircraft's kinetic energy.
Key Points:
1. Work by wire — The wire opposes motion: W = −F × s = −367500 × 100 = −36750000 J.
2. Energy equation — Change in KE = W → 0 − ½mv² = −36750000.
3. Solve for v² — v² = (2 × 36750000)/15000 = 4900.
4. Velocity — v = √4900 = 70 m s⁻¹ (about 252 km h⁻¹).
5. Principle — The work-energy theorem lets us find the speed without knowing the detailed deceleration profile.
Conclusion: The aircraft was travelling at about 70 m s⁻¹ (252 km h⁻¹) just before the arresting wire caught it.

Introduction: The force-displacement graph lets us compute work as the area, which feeds into the work-energy theorem.
Key Points:
1. Speed at 0 m — KE = ½mv² → 180 = ½ × 10 × v² → v = 6 m s⁻¹.
2. Work from graph — Work = area under the F-s graph from 0 to 4 m (add rectangular/triangular areas as per graph values).
3. KE and speed at 4 m — KE₄ = KE₀ + W; then v₄ = √(2 × KE₄ / m).
4. Negative acceleration — In any portion where the force (and hence area contribution) is negative, the block decelerates there.
Conclusion: By combining the graph-derived work with the initial KE, we can track the block's speed everywhere along the track.

Introduction: An inclined plane's geometry determines its mechanical advantage and the effort required to raise a load.
Key Points:
1. Ramp length — By Pythagoras: AC = √(30² + 40²) = √2500 = 50 cm.
2. MA calculation — MA = L/h = 50/30 = 1.67, meaning the effort is about 60% of the load weight.
3. Longer ramp — Increasing L (gentler slope) increases MA, so a smaller effort force is needed.
4. Total work — Total work done (mgh) is unchanged; a smaller force is applied over a proportionally larger distance.
Conclusion: The inclined plane trades distance for force, keeping the total work done the same regardless of ramp length.

Introduction: Energy can be transformed from one form to another in various everyday and natural processes.
Key Points:
1. Truck uphill — Kinetic energy → gravitational potential energy and heat (against friction).
2. Watch spring — Elastic potential energy stored in the spring → kinetic energy of the gears and hands.
3. Photosynthesis — Light energy from the Sun → chemical energy stored in glucose molecules in plants.
4. Water from dam — Gravitational potential energy → kinetic energy → electrical energy via a turbine.
5. Burning matchstick — Chemical energy in the match → thermal energy (heat) and light energy.
6. Solar panel and bulb — Solar panel: light energy → electrical energy; bulb: electrical energy → light and heat.
7. Microphone — Sound energy (vibrations) → electrical energy (signal).
Conclusion: In every case, energy is not created or destroyed but converted from one form to another as needed.