Where Mindful Learning Begins
Where Mindful Learning Begins
Complete NCERT Solutions for Chapter 7 of the new Class 9 Science Exploration textbook (CBSE 2026-27) — every Think It Over, Activity, Pause & Ponder, Worked Example, Ready to Go Beyond, Bridging Science and Society, Revise Reflect Refine, and Journey Beyond question on this one page, with full step-by-step working for every numerical.
This chapter reframes force and motion in terms of energy — when work is actually done (and when it isn't, despite the effort involved), how kinetic and potential energy convert into each other, and how simple machines like levers and inclined planes trade force for distance without creating extra work from nothing. The work-energy theorem, the law of conservation of energy, and the mechanical advantage/velocity ratio/efficiency formulas here are among the most frequently tested ideas in Class 9 board papers.
Work, Energy, and Simple Machines builds a single powerful idea from the ground up — that the work done on an object equals the change in its energy — and then uses this work-energy theorem to explain kinetic energy, potential energy, the conservation of mechanical energy, power, and finally simple machines like the pulley, the inclined plane, and the lever. Every question is solved here, section by section, exactly as the textbook presents them, with full working for every numerical.
What counts as work, when it is zero, positive or negative, and how work done equals the change in an object's energy.
Deriving K = ½mv² and U = mgh, and showing that mechanical energy is conserved in free fall and pendulum motion.
Rate of doing work, and how a pulley, inclined plane, or lever trades off force against distance without reducing total work.
When is work done?
| Situation | Work done? | Why |
|---|---|---|
| Force causes displacement in the same direction | Positive work | Force and displacement are aligned |
| Force acts opposite to displacement | Negative work | e.g., friction slowing a moving object |
| Force is perpendicular to displacement | Zero work | e.g., carrying a bag while walking horizontally |
| No displacement occurs | Zero work | e.g., pushing against a wall that doesn't move |
Key formulae
where \(F\) = force, \(s\) = displacement, \(\theta\) = angle between force and displacement, \(m\) = mass, \(v\) = velocity, \(g\) = acceleration due to gravity, \(h\) = height. SI unit of work and energy is the joule (J); SI unit of power is the watt (W).
Simple machines
The law of conservation of energy: energy can neither be created nor destroyed, only transformed from one form to another — the total energy of an isolated system always remains constant.
Answer: using conservation of mechanical energy, the potential energy at the top converts entirely to kinetic energy at the bottom (ignoring friction). If the slide has height \(h\):
The velocity depends only on the height \(h\) of the slide, not on the mass of the child or the shape of the slide. This is worked out in full in Example 7.8 of the chapter.
Answer: yes. From \(v = \sqrt{2gh}\), the velocity at the bottom depends only on the height \(h\) and the acceleration due to gravity \(g\) — not on the mass of the child. Therefore, two children of different masses will reach the bottom of the same slide with the same speed (assuming friction is neglected).
This is a direct consequence of the conservation of mechanical energy and the fact that \(g\) is the same for all masses — the mass \(m\) cancels out on both sides of \(mgh = \tfrac{1}{2}mv^2\).
Answer: the slide with the greatest vertical height will produce the largest velocity at the bottom, since \(v = \sqrt{2gh}\) and \(v\) increases with \(h\). The shape of the slide (straight, curved, or spiral) does not matter — only the vertical height \(h\) from top to bottom determines the final speed.
Why the ball creates a depression: when the ball falls and hits the sand, it applies a force on the sand grains, compressing them downward. The ball's kinetic energy at impact is transferred to the sand as work done in compressing it. The greater the kinetic energy, the deeper the depression.
Comparison of Depressions
Conclusion: the greater the height from which the ball is dropped, the more potential energy it has, the greater its kinetic energy at impact, and the deeper the depression in the sand. This confirms that potential energy increases with height: \(U = mgh\).
Setup: the bob starts at P with potential energy \(mgh\) and zero kinetic energy. It swings through Q (lowest point) where it has maximum kinetic energy and zero potential energy (taking Q as reference). It then swings up to R on the other side.
Observation: the bob almost reaches the same height \(h\) on the other side (point R almost equals the height of P). At R, kinetic energy is again zero and potential energy is \(mgh\).
What This Demonstrates
This is the conservation of mechanical energy — in the absence of friction and air resistance, the total mechanical energy of the pendulum remains constant.
Why the bob doesn't exactly reach the same height in real life: energy is lost due to (i) friction at the pivot support, and (ii) air resistance. The pendulum therefore slows down gradually and eventually stops at Q, the lowest point.
Conclusion: an inclined plane reduces the effort needed to raise an object to a height, at the cost of applying the force over a greater distance. The product (force × distance = work) stays the same in all cases, consistent with the work-energy theorem. The mechanical advantage = \(L/h > 1\).
Observation: a much lighter eraser (or two) placed on the longer side of the scale can lift the much heavier stapler on the shorter side.
Principle — the lever: the scale acts as a rigid bar (lever) that rotates about the fulcrum (pencil). Using Eq. (7.14):
Here \(F_1\) is the effort (lighter eraser force) with effort arm \(d_1\) (longer distance from fulcrum), and \(F_2\) is the load (heavier stapler force) with load arm \(d_2\) (shorter distance from fulcrum). Because \(d_1 > d_2\), we get \(F_1 < F_2\) — a small force applied over a large distance on the effort side produces a large force over a small distance on the load side, giving a mechanical advantage greater than 1.
A lever does not reduce the total work done — it only trades off force against distance.
Observation: the beam balances when
| Coins in left pan (n₁) | Distance L₁ | Coins in right pan (n₂) | Distance L₂ needed |
|---|---|---|---|
| 1 | L | 1 | L (balanced at centre) |
| 1 | L | 2 | L/2 |
| 1 | L | 4 | L/4 |
| 1 | L | 8 | L/8 |
Conclusion: the product (number of coins × distance from fulcrum) is always equal on both sides when the beam is balanced. This confirms Eq. (7.15) and the lever principle: effort arm × effort = load arm × load.
Answer: no. Work is defined as \(W = F \times s\) (force × displacement in the direction of force). Although the weightlifter applies an upward force on the barbell equal to its weight, the barbell does not move — its displacement is zero (\(s=0\)). Therefore, work done \(= F \times 0 = 0\) J. She does no work on the barbell in the scientific sense, even though her muscles are using energy to maintain the force.
Answer: the work done by friction on the stack of coins is negative. Friction acts in the direction opposite to the motion of the coins (backward), while the coins' displacement is forward. Since force and displacement are in opposite directions, the work done by friction is negative. This negative work reduces the kinetic energy of the coins, slowing them down and eventually bringing them to rest.
Answer: the muscular (chemical) energy is transformed into several forms as you ride:
On a flat road at constant speed, muscular energy is continuously converted mainly into heat (overcoming friction and air resistance), since kinetic energy stays constant. If the rider accelerates, more muscular energy goes into increasing kinetic energy.
The ratio of velocities \(v_A : v_B = 2 : 1\). Object A (lighter) must move twice as fast as object B (heavier) to have the same kinetic energy.
Answer: no. Kinetic energy \(K = \tfrac{1}{2}mv^2\). If the object moves with constant velocity, \(v\) is unchanged, and since mass \(m\) is also constant, \(K\) remains constant regardless of the object's position. Position has no direct effect on kinetic energy — only speed does.
Horizontal motion at constant velocity: no change in potential energy. Gravitational PE = \(mgh\), where \(h\) is the vertical height above the reference level. Horizontal movement does not change \(h\), so \(U = mgh\) remains constant.
Vertical upward motion: yes, potential energy increases. As \(h\) increases, \(U = mgh\) increases proportionally. The work done against gravity in lifting the object is stored as increased gravitational potential energy.
Just before hitting the ground (point C), height \(h' = 0\), so potential energy = 0. The ball started from height \(h\) with zero velocity. By conservation of energy, all PE has converted to KE:
Mechanical energy at C = KE + PE = \(mgh + 0 = mgh\). This equals the mechanical energy at point A (\(mgh + 0 = mgh\)), confirming that mechanical energy is conserved throughout the free fall.
This can also be verified using kinematics: \(v^2 = 2gh\), so KE \(= \tfrac{1}{2}m(2gh) = mgh\).
Why C, D, E are progressively lower: in the real world, friction between the ball and the track, and air resistance, converts some mechanical energy into thermal energy at every point. Each time the ball rises to a peak, slightly less mechanical energy is available, so the peak height is lower. Eventually all mechanical energy is lost to heat and the ball comes to rest at the lowest point. Yes — this is entirely due to energy lost to friction and air resistance. In a perfectly frictionless system, the ball would reach exactly the same height at every peak.
Answer: a road going straight up a hill would have a short length \(L\) but the full height \(h\) to climb, meaning the mechanical advantage \(= L/h\) is small. The effort required to push or pull a vehicle up would be nearly equal to the full weight component.
By winding around in gentle slopes, the road length \(L\) becomes much greater than the height \(h\) gained. From the inclined plane formula, mechanical advantage \(= L/h\). A larger \(L\) means a smaller force is needed for the same height gain. For example, a mountain road that winds 5 km to climb 200 m vertically has MA \(= 5000/200 = 25\) — vehicles need only 1/25th of the force they would need going straight up.
The total work done is the same in both cases (\(W = mgh\)), but the gentle slope spreads this work over a much longer distance with proportionally less force — well within the capabilities of vehicle engines, and safer.
Answer: an inclined ladder acts as an inclined plane. Its length \(L\) is greater than the vertical height \(h\) gained, so mechanical advantage \(= L/h > 1\) — a smaller force component is needed compared to the full weight needed for a vertical ladder.
On a vertical ladder, you must lift your full body weight with each step — the force required equals \(mg\). On an inclined ladder, part of your weight is supported by the rungs horizontally, and the force along the slope is \(mg\sin\theta\) (where \(\theta\) is the angle from horizontal), which is less than \(mg\). The effort required to climb is thus reduced, making it less tiring.
Answer: the spoon acts as a Class I lever with the rim of the can as the fulcrum. The effort is applied at the far end of the spoon (long effort arm), while the lid-opening force (load) acts at the tip of the spoon near the rim (short load arm). By Eq. (7.16), MA = effort arm ÷ load arm. Since the effort arm is much longer than the load arm, the mechanical advantage is much greater than 1 — a small downward force applied on the spoon handle generates a much larger upward force on the can lid, easily prying it open.
Answer: scissors act as a Class I lever with the pivot (screw) as the fulcrum. The cutting force (load) acts where the object is placed, and the effort is applied at the handles. When a hard object is placed closer to the fulcrum (shorter load arm), the mechanical advantage increases: MA = effort arm ÷ load arm. A shorter load arm means the force applied on the object by the blades is larger for the same handle effort — this greater cutting force can cut through the hard material. Placing the object far from the pivot (long load arm) gives less cutting force, suitable only for soft materials.
Answer: a perpetual motion machine would require a machine to continuously do useful work without any energy input — in other words, it would need to create energy from nothing. This is impossible because of two fundamental principles:
Therefore, every real machine has an efficiency less than 100%. Its mechanical energy decreases over time as it is converted to thermal energy, and it eventually slows down and stops unless new energy is supplied from an external source (fuel, electricity, etc.).
A perpetual motion machine violates the First Law of Thermodynamics (conservation of energy). This is why no design has ever worked, despite centuries of attempts.
Lifting upward: the girl applies an upward force equal to the dumbbell's weight. The dumbbell moves upward — displacement is in the same direction as the applied force, so the work done is positive. The dumbbell gains gravitational potential energy.
Lowering downward: the girl still applies an upward force to control the descent, but the dumbbell moves downward — displacement is opposite to the applied force, so the work done by the girl is negative.
The goalkeeper applies a force opposite to the ball's motion, so the displacement is taken as negative relative to that force.
The goalkeeper does −30 J of work on the ball (negative work). This negative work removes kinetic energy from the ball, bringing it to rest.
Collision 1 — Striker → White Coin
The striker applies a forward force on the white coin (in the direction of its displacement): the striker does positive work on the white coin, which gains energy. By Newton's Third Law, the white coin applies an equal backward force on the striker: the white coin does negative work on the striker, which loses energy and slows down.
Collision 2 — White Coin → Black Coin
The white coin applies a forward force on the black coin: the white coin does positive work on the black coin, which gains energy. By Newton's Third Law, the black coin applies a backward force on the white coin: the black coin does negative work on the white coin, which loses energy and slows down (or stops).
Energy flow: Striker → White coin → Black coin, via positive work done at each collision.
The kinetic energy becomes 4 times the original value when velocity doubles, because KE depends on \(v^2\) — doubling \(v\) quadruples KE.
This is why stopping distance increases by 4 times when speed doubles (Chapter 4) — the brakes must do 4 times more work to remove 4 times the kinetic energy.
Mass \(m = 0.2\) kg; velocity \(v = 154.8\) km/h \(= 154.8 \times \tfrac{1000}{3600} = 43\) m/s
Work done by wire on aircraft \(= F \times s = 367500 \times (-100) = -36{,}750{,}000\,\text{J}\) (negative because force and displacement are opposite).
By the work-energy theorem, \(W = \Delta KE\):
\(v = 70\) m/s \(= 252\) km/h.
The wire must withstand 367,500 N — nearly 37 tonnes of force — to stop the aircraft, which is why aircraft carrier arresting wires are specially engineered high-strength steel cables.
At top: PE \(= mgh\), KE \(= 0\). At bottom: PE \(= 0\), KE \(= \tfrac{1}{2}mv^2\). By conservation:
The velocity depends only on \(h\) and \(g\) — mass \(m\) cancels out, so heavier or lighter children reach the same speed. The shape of the slide also does not matter — only the vertical height difference determines the final speed.
\(v = 72\) km/h \(= 20\) m/s. Initial KE \(= \tfrac{1}{2} \times 10000 \times 400 = 2{,}000{,}000\,\text{J}\). Let ramp length \(= d\); height gained \(= d/2\) (from the hint given).
Using the work-energy theorem, \(W_{\text{sand}} = \Delta(\text{total energy})\):
The minimum ramp length is 20 m.
Work done \(= mgh = 75 \times 10 \times 2 = 1500\,\text{J}\)
\(v = 72\) km/h \(= 20\) m/s, \(u = 0\). Work done \(= \Delta KE = \tfrac{1}{2} \times 1000 \times (20)^2 - 0 = 200{,}000\,\text{J}\)
Height \(h = 30\) cm, width \(= 40\) cm. Length of ramp \(L = \sqrt{30^2 + 40^2} = \sqrt{2500} = 50\) cm (right-angled triangle property).
The ramp reduces the force required to about 60% of the object's weight, but the force must act over 50 cm instead of 30 cm.
Using the lever balance condition, load × load arm = effort × effort arm:
The 15 kg child sits at seat A (2 m from fulcrum) and the 30 kg child sits at seat D (1 m from fulcrum). The seesaw is balanced because \(15 \times 2 = 30 \times 1 = 30\) kg·m.
Answer: the work done is zero. Work = force × displacement in the direction of force. If force and displacement are perpendicular, the displacement in the direction of force is zero, so \(W = 0\). Example: a girl carries a box while walking horizontally — she applies an upward force to support the box's weight, but the box moves horizontally. Since force is vertical and displacement is horizontal (perpendicular to each other), no work is done by the carrying force on the box. In higher grades, this is expressed as \(W = Fs\cos\theta\), where \(\cos 90^\circ = 0\).
Answer: force is a vector (magnitude and direction) and displacement is also a vector. However, their combination in calculating work — \(W = Fs\cos\theta\) — gives a scalar (a single number, with only magnitude, positive or negative). Work does not point in any spatial direction. It is described as positive (force aids displacement), negative (force opposes displacement), or zero (force perpendicular to displacement). This type of product of two vectors that gives a scalar is called the scalar product or dot product, studied in higher grades.
Answer: energy can be stored by changing the relative arrangement or positions of objects in a system that interact through forces:
In all cases, the system stores potential energy because work was done against the internal forces to create the separation or arrangement.
Answer: potential energy can be stored and retrieved when the internal forces can undo the deformation or rearrangement (like a spring returning to its original shape, or magnets moving back together). Such forces are called conservative forces — gravitational, electric, and magnetic forces are conservative, and work done against them can be fully recovered.
Friction is a non-conservative (dissipative) force. Work done against friction converts mechanical energy into thermal energy (heat), which disperses into the surroundings. This energy cannot be recovered as mechanical energy — it is permanently lost to the environment. Therefore, work done against friction does not lead to stored potential energy.
Answer: in a fixed pulley, effort equals load (MA = 1) — it only changes the direction of force. In a movable pulley system, the load is attached to a pulley that can move. One end of the rope is fixed; the effort is applied at the other end. The load is supported by two segments of the rope, each carrying half the load, so effort = load/2 and MA = 2.
With more pulleys in a system (block and tackle), the rope segments supporting the load increase, and MA increases proportionally (3, 4, 5, etc.). The trade-off: the effort moves through a much larger distance than the load. This is used in cranes, elevators, and construction equipment to lift very heavy objects with small applied forces.
| Class | Arrangement | Mechanical Advantage | Examples |
|---|---|---|---|
| Class I | Fulcrum in between effort and load | Can be > 1 or < 1 | Scissors, crowbar, pliers, balance scale, seesaw, tongs |
| Class II | Load in between fulcrum and effort | Always > 1 | Lemon squeezer, wheelbarrow, bottle opener, nutcracker |
| Class III | Effort in between fulcrum and load | Always < 1 (used for speed/range, not force) | Tweezers, broom, hammer, fishing rod, oar, human forearm |
Answer: yes — the Law of Conservation of Energy is a universal principle: the total energy of an isolated system (one not acted upon by external forces) remains constant. Energy can change from one form to another — kinetic, potential, thermal, chemical, electrical, nuclear, sound, light — but the total amount is always conserved. Mechanical energy conservation is a special case, valid when only conservative forces (gravity, springs, magnetic/electric forces) do work and there is no friction or dissipation.
The watermill uses the natural energy of flowing water to do mechanical work:
Modern hydroelectric power works on the same principle at a much larger scale: water stored in a dam (high PE) is released through pipes to turbines at the base. The kinetic energy of the water spins the turbines, which drive generators to produce electricity. The energy conversions are:
The gharat is a zero-emission, renewable energy device that has been used sustainably for centuries in mountain communities.
Correct statements: (iii) and (iv).
(i) PE gained in elevator (straight vertical lift):
(ii) PE gained by climbing stairs: the staircase is a longer path, but the vertical height gained is the same (72.5 m). Therefore \(U = mgh = 36{,}250\,\text{J}\) — same as the elevator.
(iii) Conclusion: gravitational potential energy depends only on the vertical height \(h\), not on the path taken to reach that height. Whether you go straight up in an elevator or wind up a staircase, the gain in PE is the same, because gravity is a conservative force.
However, the total work done by the student's muscles is more when climbing stairs, because extra work is done against friction at each step. The PE gain is the same, but the total energy expenditure differs.
Let the height of each floor \(= H\). Height of 10th floor \(= 10H\); height of 20th floor \(= 20H\).
The crane requires 2 times (double) the energy to lift to the 20th floor.
The power required is the same for both tasks. Even though the 20th floor task requires double the energy, it also takes double the time, so the rate of work (power) is unchanged.
Factors Determining Energy (Work) Required
Does speed of raising change work done? No. Work done \(W = mgh\) is independent of how fast or slowly the flag is raised — the same energy is needed whether the flag is raised in 10 seconds or 10 minutes.
If speed doubles: Power \(= W/t\). If the flag is raised in half the time (double speed), \(P_2 = W/(t/2) = 2W/t = 2P_1\). Power doubles when speed doubles, since work stays the same but time halves.
The fuel provides kinetic energy to the scooter+rider system (no friction/air resistance losses given).
Ratio of fuel \(= KE_2/KE_1 = 100v^2/80v^2 = 5/4\).
The ratio of fuel used is 5 : 4 — the scooter uses 25% more fuel on the second day due to the increased mass.
Using the lever balance condition, load × load arm = effort × effort arm. Let child's mass \(= m\), so adult's mass \(= 2m\). For balance:
The child must sit twice as far from the fulcrum as the adult. Example: if the adult sits 1 m from the fulcrum, the child sits 2 m from the fulcrum.
Figure description: a horizontal beam with a fulcrum (triangle) in the middle. On the left, the child sits at distance \(2d\) from the fulcrum. On the right, the adult sits at distance \(d\) from the fulcrum. Both sides exert equal turning effects (moments): \(mg \times 2d = 2mg \times d\), so the seesaw stays horizontal and balanced.
(i) During upward motion: gravity acts downward, displacement is upward — opposite directions, so work by gravity is negative (\(W_g = -mgh\)). During downward motion: gravity acts downward, displacement is also downward — same direction, so work by gravity is positive (\(W_g = +mgh\)).
(ii) Without air resistance, the ball would reach: \(h_{\max} = u^2/(2g) = 400/20 = 20\) m. But it only reaches 19.4 m — a difference of 0.6 m.
Work done by air resistance \(= -12\,\text{J}\) (negative, since air resistance opposes the upward motion).
(i) Speed at 0 m: \(KE = \tfrac{1}{2}mv^2 = 180\,\text{J}\)
(ii) Work done by the force from 0 to 4 m = area under the force-displacement graph. Reading the graph: the force rises from 0 to 50 N at \(x=2\) m (triangle), then decreases back to 0 at \(x=4\) m (triangle). Total area \(= \tfrac{1}{2}\times 2\times 50 + \tfrac{1}{2}\times 2\times 50 = 100\,\text{J}\).
Negative acceleration: in the region from 0 to 4 m, the force is always in the direction of motion (never negative), so acceleration is positive (or zero) throughout — there is no region of negative acceleration.
Using conservation of energy: \(\tfrac{1}{2}mv^2 = mgh \;\Rightarrow\; h = v^2/(2g)\). The same initial velocity \(v\) is used on both Earth and Moon.
The ball will reach a height of 48 m on the Moon — six times higher than on Earth.
(i) A to B: the velocity-time graph is a horizontal line at 35 m/s from \(t=0\) to \(t=1\) s. The car moves at constant speed during this interval — the driver's reaction time before the brakes are applied. No deceleration occurs yet.
(ii) KE at A: \(v=35\) m/s, \(m=1000\) kg
(iii) At C, velocity \(=0\), so KE at C \(=0\). Work done by brakes \(= \Delta KE = 0 - 612{,}500 = -612{,}500\,\text{J}\).
(iv) The kinetic energy is converted into thermal energy (heat) — the brakes, tyres, and road become hot. Some energy is also converted to sound (screeching of tyres). None of this energy is recovered.
Since the track is frictionless, mechanical energy is conserved: KE + PE = constant \(= 30\,\text{J}\) (at O, KE = 0, PE = 30 J).
The ball cannot reach R, because its total mechanical energy (30 J) is less than the potential energy at R (40 J). The ball would turn back before reaching R.
A ball on a frictionless track can never reach a point where PE exceeds its total mechanical energy, just as a roller coaster cannot rise higher than the height from which it started at rest.
(i) Velocity just before impact (conservation of energy):
(ii) Depth of depression: KE at impact \(= \tfrac{1}{2} \times 1.5 \times 200 = 150\,\text{J}\). This KE does work against the sand's resistive force over depth \(d\):
A harder surface (larger resistive force) would create a shallower depression — wet sand has lower resistance than dry compacted sand, hence the deeper impression on wet sand beaches.
Energy conversion: elastic potential energy (stored in the stretched rubber band) converts to kinetic energy of the refill as it is launched — identical to a slingshot (gulel): the work done in stretching the band is stored as elastic PE, which converts to KE of the refill on release.
Relationship between stretch and distance: the more the rubber band is stretched, the more elastic PE is stored (PE ∝ stretch², similar to a spring where \(U = \tfrac{1}{2}kx^2\)). Greater PE means greater KE on release, greater launch velocity, and greater distance travelled (projectile motion). Therefore distance ∝ (stretch)² for an ideal elastic band — doubling the stretch should approximately quadruple the distance, though real rubber bands may deviate from this ideal relationship.
Procedure — Lever
Use a 50 cm ruler as the beam with a pencil as fulcrum at the 25 cm midpoint. Hang a paper cup on each end using thread; place known weights (coins) in one cup (load) and coins in the other (effort), adjusting position until balanced. Measure effort arm (\(L_1\)) and load arm (\(L_2\)); calculate \(MA = L_1/L_2\).
Procedure — Inclined Plane
Use a cardboard ramp at various angles. Pull a small wheeled cart using a spring balance along the ramp versus lifting it vertically. Record effort (spring balance reading) and load (weight of cart); calculate \(MA = \text{load}/\text{effort} = L/h\).
What the experiment shows: simple machines reduce the force required for a task (\(MA > 1\)) at the cost of increased distance. The product of force × distance (work) is always equal on the input and output sides (ignoring friction). Machines do not create or destroy energy — they only redistribute force and distance.
Key insight: conservation of energy is verified visually — friction always reduces mechanical energy, and the forms of energy change but the total never increases without external input.
Background: astronomers have observed that the Universe is not just expanding (as expected from the Big Bang) but its expansion is accelerating — surprising, since gravity should be pulling galaxies together and slowing the expansion.
Dark energy: to explain the accelerating expansion, scientists propose that roughly 68% of the total energy of the Universe consists of a mysterious "dark energy" that acts as a repulsive force, pushing space itself to expand faster and faster.
Why It Is Mysterious
Significance: dark energy may govern the ultimate fate of the Universe. If it continues to increase, it could eventually tear apart galaxies, stars, and even atoms in a scenario called the "Big Rip". Understanding dark energy is one of the greatest open questions in modern physics and cosmology.
Work done on an object equals the change in its energy — this single work-energy theorem explains why a moving ball can knock over a wicket, why a raised pot can shatter on impact, why kinetic and potential energy trade places as a pendulum swings while their sum stays constant, and why every simple machine, from a pulley to a lever, can only trade force for distance and never reduce the total work that must be done.
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