Comprehensive Study Guide to Work, Energy, and Simple Machines

Foundations of Work, Energy, and Motion

• Limitations of Direct Newtonian Analysis:

  • Kinematic equations and Newton's laws describe motion effectively when forces are constant and directions are straightforward.

  • When forces change with time or act in complex, varying directions, direct application of Newtonian mechanics becomes mathematically cumbersome.

  • Work, energy, and power provide an alternative scalar framework that simplifies the analysis of complex mechanical interactions.

• Fundamental Definition of Energy:

  • Energy represents the capacity to perform work.

  • Daily human and technological activities rely entirely on energy transfer from diverse sources:

  • Biological systems derive energy from food to power muscular exertion and locomotion.

  • Electrical energy drives household appliances, such as rotating a mechanical fan.

  • Chemical fuel combustion provides thermal and mechanical energy to propel automobiles.

• Role of Simple Machines:

  • Simple machines serve as fundamental mechanical devices that facilitate tasks by altering the magnitude or direction of an applied force.

  • They do not diminish the total work required to accomplish a given task, but provide convenience and reduce required operational effort.

Work Done by a Constant Force

• Physical Concept and Intuition:

  • Lifting a wheat bag of mass $m = 5\,\text{kg}$ vertically at a steady, slow pace requires applying an upward force equal to the gravitational weight $mg$, where $g$ is the acceleration due to gravity.

  • When lifted through a height of $1\,\text{m}$, the force acts in the exact direction of the displacement.

  • Work scaling rules:

  • Lifting 3 identical bags successively to $1\,\text{m}$ requires 3 times the work and consumes 3 times the fuel if performed by a machine.

  • Lifting all 3 bags simultaneously to $1\,\text{m}$ requires 3 times the upward force ($3F = 3mg$) over the same displacement, resulting in 3 times the work.

  • Lifting a single bag to a height of $3\,\text{m}$ requires applying the same force over 3 times the distance, which also triples the work done.

  • General principle: Applying a larger force over the same distance, or the same force over a larger distance, yields a proportional increase in work.

• Scientific Definition of Work:

  • The work done on an object by a constant force is defined as the product of the magnitude of the applied force and the displacement in the direction of that force:     $\text{Work done} = \text{Force applied} \times \text{Displacement in the direction of the force}$

  • Expressed algebraically for an object displaced by $s$ along the direction of a constant force $F$:     $W = F \times s$

  • Contextual specification: Whenever work is evaluated, the specific force (or agency) performing the work and the specific object on which the work is performed must be explicitly identified.

• Units and Dimensions of Work:

  • The SI unit of work is the joule ($\text{J}$), named in honour of James Prescott Joule.

  • $1\,\text{joule}$ is defined as the amount of work done when a constant force of $1\,\text{newton}$ ($\text{N}$) displaces an object through a distance of $1\,\text{metre}$ ($\text{m}$) along the direction of the force:     $1\,\text{J} = 1\,\text{N} \times 1\,\text{m}$

  • Dimensional derivation using base SI units:     $1\,\text{N} = 1\,\text{kg}\,\text{m}\,\text{s}^{-2}$     $1\,\text{J} = (1\,\text{kg}\,\text{m}\,\text{s}^{-2}) \times (1\,\text{m}) = 1\,\text{kg}\,\text{m}^2\,\text{s}^{-2}$

• Graphical Interpretation of Work:

  • On a graph plotting force ($y$-axis) against displacement in the direction of the force ($x$-axis), the work done corresponds directly to the area under the curve between the initial and final positions.

  • For a constant force of $10\,\text{N}$ acting across a displacement of $1.0\,\text{m}$, the work corresponds to the rectangular area:     $\text{Area} = 10\,\text{N} \times 1.0\,\text{m} = 10\,\text{J}$

  • This geometric property holds universally: even when an applied force varies continuously, the total work done is given by the integral area bounded by the force-displacement curve.

• Conditions for Zero Work:

  • Zero Force ($F = 0$): If no net force acts on the body, $W = 0$.

  • Zero Displacement ($s = 0$): If the force fails to produce motion, no work is done on the object.

  • Example: Pushing against a rigid, stationary brick wall produces zero displacement ($s = 0$), hence the mechanical work done on the wall is zero ($W = 0$).

  • Biological fatigue vs. mechanical work: A person pushing a wall still feels fatigued because internal muscle fibres undergo continuous cycles of contraction and relaxation, consuming internal chemical energy without delivering mechanical work to the external structure.

  • Perpendicular Alignment: If the force acts perpendicular to the direction of motion, the component of displacement along the force is zero, yielding $W = 0$.

  • Example: Carrying a box horizontally across a flat floor. The carrier exerts an upward vertical force to counteract gravitational weight, while the displacement is entirely horizontal. Because the angle between force and displacement is $90^\circ$, this upward supporting force performs zero work on the box.

• Sign Convention: Positive vs. Negative Work:

  • Work is a scalar physical quantity possessing magnitude and sign, but no spatial vector direction.

  • Positive Work ($W > 0$): Occurs when the applied force and the displacement act in the identical direction.

  • Example: Pushing a wheelchair forward causes forward displacement, resulting in positive work done on the wheelchair.

• Negative Work ($W < 0$): Occurs when the applied force opposes the direction of displacement.

  • Example: A soccer goalkeeper exerting a backward resisting force on an incoming ball to bring it to rest. The ball moves forward while the goalkeeper pushes backward, meaning the goalkeeper performs negative work on the ball.

  • Action-reaction dynamics: In accordance with Newton's third law, the ball exerts an equal and opposite forward force on the goalkeeper's hands while they yield backward, meaning the ball performs positive work on the goalkeeper.

  • Worked Examples on Work Done:

• Example 7.1 (Lifting and Lowering a Dumbbell):

  • Scenario: A girl lifts a dumbbell vertically and then slowly lowers it back down.

  • Upward Motion: The girl applies an upward force equal to the dumbbell's weight. Displacement is directed upward. Because force and displacement share the same direction, she does positive work on the dumbbell.

  • Downward Motion: To prevent the dumbbell from accelerating downward under free fall, she maintains an upward resisting force while displacement occurs downward. Because force and displacement are oppositely directed, she does negative work on the dumbbell.

• Example 7.2 (Goalkeeper Stopping a Ball):

  • Parameters: Applied resistive force $F = 200\,\text{N}$; displacement during braking $s = 15\,\text{cm} = 0.15\,\text{m}$.

  • Calculation: Because the ball moves opposite to the braking force, displacement relative to the force vector is negative ($s = -0.15\,\text{m}$):       $W = F \times s = 200\,\text{N} \times (-0.15\,\text{m}) = -30\,\text{J}$

  • Conclusion: The goalkeeper performs $-30\,\text{J}$ of work on the ball.

The Work-Energy Theorem

• Mechanics of Energy Transfer:

  • Performing positive work on an object transfers energy to it, establishing a capacity to do work on subsequent systems.

  • Examples:

  • A cricket fielder performs work on a ball during a throw. The airborne ball stores this energy as motion and delivers work against the wickets upon collision, knocking them down.

  • Lifting a flowerpot to a balcony requires work against gravity. Upon falling, the flowerpot releases this accumulated capacity, damaging objects below.

  • Newton's Third Law governs collision interactions: in any impact, the colliding body exerts a force in the direction of motion of the target (positive work, energy transferred to target), while the target exerts an opposing reaction force on the collider (negative work, energy depleted from collider).

• Formal Statement of the Work-Energy Theorem:

  • The net work done on an object or system is strictly equal to the net change in its energy:     $\text{Work done on an object} = \text{Change in its energy}$     $W = \Delta E$

  • This theorem remains valid universally for constant forces, continuously variable forces, and multi-body interacting systems.

  • The SI unit of energy is identical to that of work: the joule ($\text{J}$).

• Scientific Biography: James Prescott Joule (1818–1889):

  • British physicist who investigated the mechanical equivalent of heat.

  • Established that mechanical work and thermal energy are interconvertible forms of a singular, unified physical quantity, forming the basis for the modern concept of energy conservation.

• Non-Mechanical Pathways of Energy Transfer:

  • Performing mechanical work via force-induced displacement is only one method of energy transfer.

  • Alternative transfer channels include:

  • Thermal Conduction and Convection: Energy transfer driven spontaneously by a temperature differential from hotter to colder regions.

  • Electromagnetic Radiation: Direct radiative transfer without requiring material contact (e.g., solar energy propagating through the vacuum of space to reach Earth).

  • Electric Circuits: Directed transport of energy via charge carriers driven by potential differences.

  • Acoustic Waves: Mechanical energy propagation through vibrating fluid or solid particles.

  • Nuclear Reactions: Subatomic mass-energy conversion fueling stellar cores.

• Worked Example 7.3 (Collision Dynamics in Carrom):

  • Scenario: A striker hits a white coin, which subsequently strikes a black coin toward a corner pocket.

• Collision 1 (Striker and White Coin):

  • The striker exerts a forward force along the displacement of the white coin, performing positive work on the white coin and increasing its energy.

  • By Newton's third law, the white coin applies an equal, opposing force on the striker, performing negative work on the striker and reducing its kinetic energy.

• Collision 2 (White Coin and Black Coin):

  • The moving white coin exerts a forward force along the displacement of the black coin, doing positive work and accelerating it toward the pocket.

  • The black coin exerts an opposing reaction force on the white coin, doing negative work and reducing the white coin's remaining energy.

Forms and Manifestations of Energy

• Taxonomy of Energy Forms:

  • Mechanical Energy: Energy associated with macroscopic motion or spatial position.

  • Thermal Energy: Microscopic kinetic energy of molecular agitation causing sensation of warmth or elevated temperature.

  • Light Energy: Electromagnetic radiation visible to the human eye, enabling sight.

  • Sound Energy: Mechanical vibrations and compressions propagating through air or other media.

  • Electrical Energy: Potential and kinetic energy linked to the position, flow, and configuration of electric charges.

  • Nuclear Energy: Potential energy bound within atomic nuclei released via fission or fusion.

  • Chemical Energy: Potential energy residing in chemical molecular bonds within food, fuels, and batteries.

• Interconversion Between Forms:

  • Energy transforms dynamically across different physical states:

  • Electric bulbs convert electrical energy into light and thermal radiation.

  • Electric water heaters convert electrical energy into thermal energy.

  • Metabolism converts chemical energy from food into mechanical muscular output and metabolic heat.

  • A striking acoustic bell converts mechanical kinetic energy into radiating sound energy.

Mechanical Energy: Kinetic Energy

• Conceptual Definition:

  • Kinetic energy ($K$) is the mechanical energy possessed by an object strictly due to its state of motion.

  • Any moving body (a bicycle on a road, a rolling bowling ball, an aircraft in flight) possesses kinetic energy.

  • An object at rest is defined to possess zero kinetic energy ($K = 0$).

• Mathematical Derivation of the Kinetic Energy Formula:

  • Consider a rigid body of mass $m$ moving with an initial velocity $u$. A constant net force $F$ acts along the displacement $s$, producing a uniform acceleration $a$.

  • From the third equation of kinematics:     $v^2 = u^2 + 2as$

  • Solving for displacement $s$:     $s = \frac{v^2 - u^2}{2a}$

  • By Newton's second law, $F = ma$. The mechanical work done by the force is:     $W = F \times s = (ma) \times \left(\frac{v^2 - u^2}{2a}\right)$

  • Cancelling acceleration $a$ yields:     $W = \frac{1}{2}m(v^2 - u^2)$

  • By the work-energy theorem ($W = \Delta K = K_f - K_i$), if the object starts from rest ($u = 0$), its final acquired kinetic energy is:     $K = \frac{1}{2}mv^2$

• Key Characteristics of Kinetic Energy:

  • Kinetic energy is a scalar quantity and has no directional vector properties.

  • It depends linearly on mass $m$ and quadratically on speed $v$.

  • Velocity-Work relationships:

  • Positive work ($W > 0$) increases velocity and increases kinetic energy.

  • Negative work ($W < 0$) decreases velocity and diminishes kinetic energy.

  • Zero net work ($W = 0$) maintains a constant velocity and constant kinetic energy.

• Worked Examples on Kinetic Energy:

  • Example 7.4 (Effect of Doubling Velocity):

  • Initial State: $K_1 = \frac{1}{2}mv^2$.

  • New State: Velocity becomes $v' = 2v$.

  • Calculation: $K_2 = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 \times \left(\frac{1}{2}mv^2\right) = 4K_1$.

  • Conclusion: Doubling the velocity quadruples the kinetic energy.

  • Example 7.5 (Kinetic Energy of a Fast Cricket Delivery):

  • Parameters: Mass $m = 0.2\,\text{kg}$; speed $v = 154.8\,\text{km}\,\text{h}^{-1}$.

  • Unit Conversion: $v = 154.8 \times \frac{1000\,\text{m}}{3600\,\text{s}} = 43\,\text{m}\,\text{s}^{-1}$.

  • Calculation:       $K = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.2\,\text{kg} \times (43\,\text{m}\,\text{s}^{-1})^2 = 0.1 \times 1849 = 184.9\,\text{J}$

  • Example 7.6 (Arresting a Jet Landing on an Aircraft Carrier):

  • Parameters: Aircraft mass $m = 15000\,\text{kg}$; stopping distance $s = 100\,\text{m}$; resisting wire force $F = 367500\,\text{N}$; final velocity $v_f = 0$.

  • Work Done by Wire: Because the force opposes displacement:       $W = F \times (-s) = 367500\,\text{N} \times (-100\,\text{m}) = -36750000\,\text{J}$

  • Work-Energy Equivalence:       $\Delta K = K_f - K_i = 0 - \frac{1}{2}mv^2 = -\frac{1}{2}(15000\,\text{kg})v^2 = -7500v^2$       $-7500v^2 = -36750000$       $v^2 = \frac{36750000}{7500} = 4900\,\text{m}^2\,\text{s}^{-2}$       $v = \sqrt{4900} = 70\,\text{m}\,\text{s}^{-1}$

  • Speed in $\text{km}\,\text{h}^{-1}$:       $v = 70 \times \frac{3600}{1000} = 252\,\text{km}\,\text{h}^{-1}$

  • Conclusion: The approach velocity was $70\,\text{m}\,\text{s}^{-1}$ ($252\,\text{km}\,\text{h}^{-1}$).

Mechanical Energy: Potential Energy

• Conceptual Definition:

  • Potential energy ($U$) is the energy stored within an object or system due to its configuration, physical deformation, or the relative spatial positions of its interacting parts.

• Mechanisms of Elastic Deformation Storage:

  • Deformed Elastic Media: Pulling a slingshot (gulel), bending an archery bow, or stretching a rubber band requires an external force to overcome internal molecular restoring forces.

  • Work performed by the external force is stored as internal elastic potential energy.

  • When released, these internal restoring forces accelerate projectiles (stones, arrows, paper pellets), converting stored potential energy into kinetic energy.

  • Coiled Springs:

  • Compressing or extending a spring deforms its lattice structure against internal restoring forces.

  • Upon release, the spring restores its equilibrium configuration, exerting force on an adjacent mass and converting stored potential energy into kinetic energy.

• Potential Energy in Interacting Field Systems:

  • Magnetic Fields: Separating two attracting opposite magnetic poles ($\text{N}$ and $\text{S}$) requires work against magnetic attractive forces. When released, the poles accelerate toward one another, transforming stored magnetic potential energy into kinetic energy capable of crushing objects placed between them.

  • Electrostatic Fields: Separating opposite electric charges ($+$ and $-$) stores electrostatic potential energy within the electric field configuration.

  • Gravitational Fields: Separating an object from the Earth by lifting it upward requires work against mutual gravitational attraction. Upon release, the Earth and the object accelerate toward their common centre of mass (though Earth's motion is imperceptible due to its immense mass), converting potential energy into kinetic energy.

• Gravitational Potential Energy:

  • Experimental Exploration (Sand Bed Depressions):

  • Dropping a heavy sphere into loose sand from a height of $1\,\text{m}$ produces an impact crater.

  • Dropping the identical sphere from an elevated height of $2\,\text{m}$ produces a noticeably deeper crater.

  • Raising an object higher requires greater work against gravity, imbuing it with greater stored energy that performs greater mechanical work on the sand bed.

  • Derivation of Formula near Earth's Surface:

  • Let an object of mass $m$ rest at ground level, defined as the reference level where $U = 0$.

  • To raise the object slowly and steadily without acceleration to a vertical height $h$, an applied upward force must exactly counterbalance gravity: $F = mg$.

  • Work done by the applied force:       $W = F \times h = mgh$

  • By the work-energy theorem, this external work equals the gain in gravitational potential energy:       $U = mgh$

  • SI Unit: Joule ($\text{J}$).

  • Physical Caveats and Force Classifications:

  • Local Validity: The formula $U = mgh$ is strictly valid near the surface of the Earth, where gravitational acceleration $g$ can be treated as constant. At astronomical altitudes, $g$ decreases inversely with the square of the distance from Earth's centre.

  • Conservative vs Non-Conservative Forces: Work performed against internal conservative forces (gravitational, electrostatic, magnetic) is completely recoverable as stored potential energy. Work done against non-conservative dissipative forces (such as friction or drag) is converted irreversibly into disordered thermal energy and cannot be stored as potential energy.

• Worked Example 7.7 (Gravitational Potential Energy of a Cricket Ball):

  • Parameters: Ball mass $m = 200\,\text{g} = 0.2\,\text{kg}$; peak height $h = 10\,\text{m}$; gravitational acceleration $g = 10\,\text{m}\,\text{s}^{-2}$.

  • Calculation:     $U = mgh = 0.2\,\text{kg} \times 10\,\text{m}\,\text{s}^{-2} \times 10\,\text{m} = 20\,\text{J}$

Conservation of Mechanical Energy

• Principle of Conservation:

  • Total mechanical energy ($E$) is the sum of kinetic energy ($K$) and potential energy ($U$):     $E = K + U$

  • In an isolated system subject exclusively to conservative internal forces (such as gravity), the total mechanical energy remains strictly constant over time:     $E = K + U = \text{constant}$

• Analytical Derivation for Free Fall:

• State at Point A (Initial Position, Height $h$, Time $t = 0$):

  • Initial velocity $u = 0$.

  • Potential energy: $U_A = mgh$.

  • Kinetic energy: $K_A = \frac{1}{2}m(0)^2 = 0$.

  • Total mechanical energy:       $E_A = K_A + U_A = 0 + mgh = mgh$

• State at Point B (Intermediate Position after Falling for Time $t$):

  • Velocity acquired under uniform acceleration $g$:       $v = u + gt = 0 + gt = gt$

  • Distance fallen during time $t$:       $s = ut + \frac{1}{2}gt^2 = 0 + \frac{1}{2}gt^2 = \frac{1}{2}gt^2$

  • Remaining elevation above ground:       $h' = h - s = h - \frac{1}{2}gt^2$

  • Potential energy at Point B:       $U_B = mgh' = mg\left(h - \frac{1}{2}gt^2\right) = mgh - \frac{1}{2}mg^2t^2$

  • Kinetic energy at Point B:       $K_B = \frac{1}{2}mv^2 = \frac{1}{2}m(gt)^2 = \frac{1}{2}mg^2t^2$

  • Total mechanical energy at Point B:       $E_B = K_B + U_B = \left(\frac{1}{2}mg^2t^2\right) + \left(mgh - \frac{1}{2}mg^2t^2\right) = mgh$

• State at Point C (Impact at Ground Level, Height $h = 0$):

  • Using kinematic relation $v^2 = u^2 + 2gs = 0 + 2gh = 2gh$.

  • Potential energy: $U_C = mg(0) = 0$.

  • Kinetic energy: $K_C = \frac{1}{2}mv^2 = \frac{1}{2}m(2gh) = mgh$.

  • Total mechanical energy:       $E_C = K_C + U_C = mgh + 0 = mgh$

• Deduction: At every point during the descent, the exact loss in potential energy ($\frac{1}{2}mg^2t^2$) equals the gain in kinetic energy ($\frac{1}{2}mg^2t^2$), preserving total mechanical energy at $mgh$.

  • Demonstration via Simple Pendulum:

• A pendulum bob suspended from a fixed string oscillates between two extreme points ($P$ and $R$) through an equilibrium low point ($Q$).

• Points P and R (Extremes): The bob comes momentarily to rest ($v = 0$). Kinetic energy is zero ($K = 0$); energy exists entirely as gravitational potential energy ($U = mgh$).

• Point Q (Lowest Equilibrium Point): Elevation is zero ($h = 0$). Potential energy is zero ($U = 0$); energy exists entirely as maximum kinetic energy ($K = \frac{1}{2}mv^2 = mgh$).

• Intermediate States: Energy continuously cycles between potential and kinetic forms, maintaining a constant sum.

• Damping in Real Systems: Real pendulums gradually lose amplitude and eventually stop due to non-conservative air resistance and pivot friction, which dissipate mechanical energy into ambient thermal energy.

  • Worked Examples on Energy Conservation:

• Example 7.8 (Child on a Playground Slide):

  • Problem: Determine the velocity of a child reaching the bottom of a frictionless slide of vertical height $h$.

  • Analysis: At the top, $E = mgh$. At the bottom, $E = \frac{1}{2}mv^2$.

  • Equating: $\frac{1}{2}mv^2 = mgh \implies v^2 = 2gh \implies v = \sqrt{2gh}$.

  • Significance: The velocity depends exclusively on vertical height $h$ and gravitational acceleration $g$. It is independent of the child's mass $m$ and the geometric curvature or path of the slide.

• Example 7.9 (Runaway Truck on Highway Escape Ramp):

  • Parameters: Truck mass $m = 10000\,\text{kg}$; initial speed $v = 72\,\text{km}\,\text{h}^{-1} = 20\,\text{m}\,\text{s}^{-1}$; ramp inclination $30^\circ$ (vertical rise $h = \frac{d}{2}$, where $d$ is travel distance along the ramp); sand resistive force $F_r = 50000\,\text{N}$; final velocity $v_f = 0$; $g = 10\,\text{m}\,\text{s}^{-2}$.

  • Initial Energies: $K_i = \frac{1}{2}(10000)(20)^2 = 2000000\,\text{J}$; $U_i = 0$.

  • Final Energies: $K_f = 0$; $U_f = mgh = (10000)(10)\left(\frac{d}{2}\right) = 50000d$.

  • Energy Change: $\Delta E = E_f - E_i = 50000d - 2000000\,\text{J}$.

  • Work Done by Non-Conservative Force (Sand Resistance): $W_{\text{sand}} = -F_r \times d = -50000d$.

  • Applying Work-Energy Relation ($W_{\text{sand}} = \Delta E$):       $-50000d = 50000d - 2000000$       $100000d = 2000000$       $d = \frac{2000000}{100000} = 20\,\text{m}$

  • Conclusion: The ramp must be at least $20\,\text{m}$ long to stop the truck.

Power: The Rate of Doing Work

• Conceptual and Mathematical Definition:

  • Power measures the temporal rate at which work is performed or energy is converted.

  • Average power ($P$) is calculated as the work done ($W$) divided by the elapsed time ($t$):     $P = \frac{W}{t}$

  • Time dependence: Running up a staircase in $1\,\text{minute}$ requires substantially higher power than walking up the identical staircase in $5\,\text{minutes}$, although both actions accomplish the identical mechanical work.

• Units of Power:

  • SI Unit: The watt ($\text{W}$), named after James Watt, who developed key improvements to the steam engine.

  • $1\,\text{watt}$ represents an energy expenditure rate of $1\,\text{joule}$ per second:     $1\,\text{W} = 1\,\text{J}\,\text{s}^{-1}$

  • Horsepower ($\text{hp}$): A traditional engineering unit established historically by comparing early steam engine performance against draft horses used for stagecoaches and mine haulage:     $1\,\text{hp} = 746\,\text{W}$

• Worked Examples on Power:

  • Example 7.10 (Power of a Weightlifter):

  • Parameters: Mass lifted $m = 75\,\text{kg}$; vertical height $h = 2\,\text{m}$; time taken $t = 5\,\text{s}$; $g = 10\,\text{m}\,\text{s}^{-2}$.

  • Work Done: $W = mgh = 75\,\text{kg} \times 10\,\text{m}\,\text{s}^{-2} \times 2\,\text{m} = 1500\,\text{J}$.

  • Power Calculation: $P = \frac{W}{t} = \frac{1500\,\text{J}}{5\,\text{s}} = 300\,\text{W}$.

  • Example 7.11 (Engine Power for Car Acceleration):

  • Parameters: Car mass $m = 1000\,\text{kg}$; initial velocity $u = 0$; final velocity $v = 72\,\text{km}\,\text{h}^{-1} = 20\,\text{m}\,\text{s}^{-1}$; acceleration time interval $t = 10\,\text{s}$.

  • Work Required (Kinetic Energy Change):       $W = \Delta K = \frac{1}{2}mv^2 - 0 = \frac{1}{2}(1000\,\text{kg})(20\,\text{m}\,\text{s}^{-1})^2 = 500 \times 400 = 200000\,\text{J}$

  • Engine Power Rating:       $P = \frac{W}{t} = \frac{200000\,\text{J}}{10\,\text{s}} = 20000\,\text{W} = 20\,\text{kW}$

Principles and Mechanics of Simple Machines

• Mechanical Principles:

  • Simple machines alter the magnitude or direction of an applied input force without altering the total work demanded by conservation of energy.

  • Fundamental trade-off: A machine may reduce the necessary applied force, but the user must exert that smaller force across a proportionally larger distance ($W = F_{\text{small}} \times d_{\text{large}}$).

  • Core terminology:

  • Effort: The external input force applied to the machine.

  • Load: The resisting force or weight to be overcome by the machine.

  • Mechanical Advantage (MA): The ratio of the load to the effort:       $\text{Mechanical Advantage} = \frac{\text{Load}}{\text{Effort}}$

• Pulleys:

  • A pulley consists of a wheel mounted on an axle with a grooved rim that guides a cord, rope, or cable.

  • Fixed Pulley:

  • Fixed to a rigid overhead support.

  • Does not alter force magnitude ($\text{Effort} = \text{Load}$, yielding $\text{MA} = 1$).

  • Function: Re-orients effort direction, allowing a person to pull downward (aided by body weight) to elevate a load upward.

  • Movable Pulleys and Compound Systems:

  • The load attaches directly to a traveling pulley supported by rope segments.

  • Allows distribution of load across multiple supporting strands, achieving $\text{MA} > 1$.

  • Enables substantial loads to be hoisted using minimal input effort; heavily employed in elevators, tower cranes, and maritime rigging.

• Inclined Planes:

  • A sloping surface bridging two different elevations.

  • Experimental Verification (Spring Balance Cart): Pulling a cart vertically records a tension equal to its full weight ($mg$). Pulling the same cart up an inclined ramp records significantly lower tension on the spring balance. As the slope angle decreases (ramp becomes less steep), the required pulling force declines progressively, while the required displacement along the ramp increases.

  • Derivation of Mechanical Advantage:

  • Let an object of mass $m$ (Load = $mg$) be moved at constant speed up a frictionless incline of length $L$ to achieve vertical height $h$.

  • Work input by effort force F'$': W_{\text{input}} = F' \times L.\n * Potential energy gain: W_{\text{output}} = mgh.\n * By work conservation: F' \times L = mgh \implies \frac{mg}{F'} = \frac{L}{h}.\n * Substituting Load and Effort into the mechanical advantage formula:\n      \text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{mg}{F'} = \frac{L}{h}\n * Because length L$always exceeds height$h$,$\text{MA} > 1.\n * **Worked Example 7.12 (Ramp Over a Step):**\n * *Parameters:* Step height h = 30\,\text{cm}$; horizontal width$b = 40\,\text{cm}.\n * *Ramp Length via Pythagorean Theorem:* L = \sqrt{h^2 + b^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = 50\,\text{cm}.\n * *Mechanical Advantage Calculation:*\n      \text{MA} = \frac{L}{h} = \frac{50\,\text{cm}}{30\,\text{cm}} = \frac{5}{3} \approx 1.67\n\n* **Levers:**\n * A rigid beam or bar pivoted about a fixed axis or point called a **fulcrum**.\n * Structural Anatomy:\n * **Fulcrum:** The fixed pivot point around which rotation occurs.\n * **Load:** The resistance force to be displaced.\n * **Effort:** The external force applied to initiate action.\n * **Load Arm:** Perpendicular distance separating the load from the fulcrum.\n * **Effort Arm:** Perpendicular distance separating the effort from the fulcrum.\n * **Mechanical Principle of Moments:**\n * Work input on one arm transfers to the opposing arm: F_1 \times d_1 = F_2 \times d_2.\n * Static equilibrium condition (Principle of the Lever):\n      \text{Effort} \times \text{Effort Arm} = \text{Load} \times \text{Load Arm}\n      n_1 \times L_1 = n_2 \times L_2\n * Formulating Mechanical Advantage:\n      \text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{\text{Effort Arm}}{\text{Load Arm}} = \frac{L_1}{L_2}\n * Lengthening the effort arm relative to the load arm produces \text{MA} > 1, magnifying output force.\n * **Worked Example 7.13 (Balancing a Seesaw):**\n * *Parameters:* Symmetrical seesaw with fulcrum at C$. Seat distances:$AC = EC = 2\,\text{m}$;$BC = DC = 1\,\text{m}$. Child 1 mass$m_1 = 15\,\text{kg}$seated at$A$. Child 2 mass$m_2 = 30\,\text{kg}.\n * *Equilibrium Equation:*\n      m_1 \times AC = m_2 \times L_2\n      15\,\text{kg} \times 2\,\text{m} = 30\,\text{kg} \times L_2\n      30 = 30 L_2 \implies L_2 = 1\,\text{m}\n * *Conclusion:* Child 2 must sit at seat D$, exactly$1\,\text{m}$from fulcrum$C.\n * **The Three Classes of Levers:**\n\n| Class | Spatial Arrangement | Distinguishing Feature | Real-World Examples |\n| :--- | :--- | :--- | :--- |\n| **Class I** | Effort — **Fulcrum** — Load | Fulcrum resides centrally between effort and load. Can amplify force, amplify distance, or redirect force. | Tongs, scissors, crowbar, pliers, beam balance scale, seesaw |\n| **Class II** | Fulcrum — **Load** — Effort | Load resides centrally between fulcrum and effort. Effort arm always exceeds load arm (\text{MA} > 1); always amplifies force. | Lemon squeezer, wheelbarrow, bottle opener |\n| **Class III** | Fulcrum — **Effort** — Load | Effort resides centrally between fulcrum and load. Load arm exceeds effort arm (\text{MA} < 1); amplifies speed and displacement rather than force. | Tongs, tweezers, broom, hammer, oar |\n\n* **Compound Machines and Perpetual Motion:**\n * Modern machinery consists of integrated combinations of simple machines (pulleys, levers, inclined planes/screws).\n * **Impossibility of Perpetual Motion:** Historic concepts proposing perpetual motion machines (using unbalanced wheels, gravity shifts, or permanent magnets to generate self-sustaining work without continuous energy input) fail because all real systems encounter non-conservative friction and air resistance, dissipating mechanical energy as heat.\n\n# Practical and Societal Applications: Energy Harnessing\n\n* **Traditional Hydro-Mechanical Milling (Gharat / Panchakki):**\n * Developed across the Himalayan valleys to process grain without combustion or electricity.\n * Mechanism of Action:\n * High-altitude stream water possesses gravitational potential energy.\n * Water channels into a steep chute or flume (Pipe A), converting potential energy into high-velocity kinetic energy.\n * High-velocity water jets hit the curved blades of a horizontal submerged turbine wheel (Turbine B), driving rotational kinetic energy.\n * The rotating central axle drives an upper heavy grinding millstone (Millstone C) resting above a stationary stone bed, milling grain.\n * Modern analogue: Contemporary hydroelectric power plants operate on identical physical principles, where reservoir gravitational potential energy transforms into kinetic energy to spin electromagnetic generators.\n\n* **Cosmological Frontiers: Dark Energy:**\n * Observational astrophysics confirms that universal cosmic expansion is currently accelerating rather than decelerating under mutual gravitational attraction.\n * To explain this cosmic acceleration, theoretical physics identifies a form of energy designated as **dark energy**.\n * Dark energy does not interchange energy with traditional mechanical, electromagnetic, or chemical channels, and its cosmological dynamics determine the ultimate fate of the universe.\n\n# Questions, Explorations, and Problem Solving\n\n* **Exploratory Concept Questions ("Think It Over"):**\n * *What will be the magnitude of velocity of the child at the bottom of the blue slide?*\n * Neglecting friction, conservation of mechanical energy gives \frac{1}{2}mv^2 = mgh$. The velocity is$v = \sqrt{2gh}.\n * *Will two children of different masses reach the bottom of the same slide with the same velocity?*\n * Yes. In v = \sqrt{2gh}, mass cancels out entirely; velocity is independent of mass.\n * *Which of the slides will result in the largest magnitude of velocity for the child at its bottom?*\n * All slides sharing the identical vertical drop height h$yield identical speeds at the bottom ($v = \sqrt{2gh}), provided frictional losses are negligible, regardless of slide curvature or ramp profile.\n\n* **Conceptual Probes ("Pause and Ponder"):**\n * *Is a weightlifter holding a barbell steady overhead doing any work on the barbell?*\n * No. Because the barbell undergoes zero displacement (s = 0$), the mechanical work done on the barbell is zero ($W = 0).\n * *Is the work done by friction on a stack of coins sliding across a rough surface positive, negative, or zero?*\n * Negative. Kinetic friction opposes the forward displacement of the sliding coins, producing negative work (W = -f_k s) and diminishing their kinetic energy.\n * *When pedaling a bicycle on a flat road, in what forms does muscular chemical energy appear?*\n * Muscular chemical energy transforms into mechanical kinetic energy of the bicycle and rider, thermal energy within flexing tires and mechanical bearings, sound energy from mechanical chain drive and tire contact, and thermal energy within human muscles.\n * *Two objects A and B of mass m$and$4m possess identical kinetic energy. What is the ratio of their velocity magnitudes?*\n * K_A = K_B \implies \frac{1}{2}m v_A^2 = \frac{1}{2}(4m) v_B^2 \implies v_A^2 = 4v_B^2 \implies \frac{v_A}{v_B} = \frac{2}{1}.\n * *Does the kinetic energy of an object moving with constant velocity change with its position?*\n * No. Kinetic energy depends strictly on mass and speed (K = \frac{1}{2}mv^2). If velocity is constant, kinetic energy remains unchanged regardless of spatial position.\n * *Does the potential energy of an object near Earth's surface change if it moves with constant velocity horizontally? What if raised vertically?*\n * Horizontal motion maintains constant vertical elevation (h = \text{constant}$), so potential energy ($U = mgh) remains constant.\n * Raising the object vertically increases elevation (h), increasing gravitational potential energy.\n * *For an object in free fall, what is its mechanical energy immediately before ground impact?*\n * Immediately prior to impact, height h = 0$, so$U = 0$. Velocity satisfies$v^2 = 2gh$, so$K = \frac{1}{2}m(2gh) = mgh$. Total mechanical energy remains$E = K + U = mgh.\n * *In a science park marble roller coaster, why do subsequent loops and crests have lower heights?*\n * Rolling friction between marble and track, along with aerodynamic drag, continually converts mechanical energy into non-recoverable thermal energy. Each subsequent crest must be lower because the total remaining mechanical energy is depleted.\n * *Why are mountain roads engineered to wind around gentle slopes rather than climbing straight up?*\n * A winding switchback acts as an inclined plane, increasing travel distance L$to attain vertical height$h$. Increasing$L$reduces the effort force required from vehicles to overcome gravity ($F = \frac{mgh}{L}).\n * *Why is climbing an inclined ladder easier than climbing a vertical ladder?*\n * An inclined ladder functions as an inclined plane. A portion of the climber's weight is supported normal to the inclined surface, reducing the parallel effort required per step.\n * *Why is it easier to pry open a can lid using a spoon?*\n * The spoon acts as a Class I lever. The rim of the can acts as a fulcrum. Applying effort at the long spoon handle provides an effort arm substantially longer than the load arm beneath the lid, multiplying the force exerted on the lid (\text{MA} > 1).\n * *Why do we position hard materials close to the scissor hinge when cutting?*\n * Moving the object closer to the pivot (fulcrum) reduces the load arm. Because \text{MA} = \frac{\text{Effort Arm}}{\text{Load Arm}}, reducing the load arm increases mechanical advantage, exerting greater cutting force on the material.\n * *Why do all real machines eventually slow down and stop?*\n * Internal and external friction, acoustic dissipation, and air resistance continuously perform negative work against mechanical components, dissipating mechanical energy as ambient thermal energy.\n\n* **Comprehensive Exercise Solutions:**\n * **Exercise 1 (True or False):**\n * (i) *Work is said to be done when a force is applied, even if the object does not move.* — **False** (s = 0 \implies W = 0).\n * (ii) *Lifting a bucket vertically upward results in positive work done on the bucket.* — **True** (applied lifting force and vertical displacement share the same upward direction).\n * (iii) *The SI unit for both work and energy is joule (\text{J}).* — **True**.\n * (iv) *A motionless stretched rubber band has kinetic energy.* — **False** (a motionless body has zero kinetic energy; it stores elastic potential energy).\n * (v) *Energy can change from one form to another.* — **True**.\n * **Exercise 2 (Fill in the Blanks):**\n * (i) \text{Work done} = \mathbf{\text{force}} \times \mathbf{\text{displacement}}\,(\text{in the direction of force}).\n * (ii) 1\,\text{joule of work is done when a force of } \mathbf{1}\,\text{newton displaces an object by } 1\,\text{metre in the direction of the force}.\n * (iii) \text{The expression for kinetic energy of a body of mass } m \text{ and velocity } v \text{ is } \mathbf{\frac{1}{2}mv^2}.\n * (iv) \text{The potential energy of an object of mass } m \text{ at a small height } h \text{ from the Earth's surface is } \mathbf{mgh}.\n * (v) \text{Power is defined as the } \mathbf{\text{rate}} \text{ at which work is done}.\n * **Exercise 3 (Ball Thrown Vertically at Apex):**\n * *Options:* (i) Force acting is zero; (ii) Acceleration is zero; (iii) Kinetic energy is zero; (iv) Potential energy is maximum.\n * *Correct Statements:* **(iii)** and **(iv)**. At the peak, instantaneous velocity is zero (v = 0 \implies K = 0$) and height is at its maximum ($h = h_{\text{max}} \implies U = \text{maximum}$). Gravity and downward acceleration ($g) continue acting unchanged.\n * **Exercise 4 (Energy Transformations):**\n * (i) *Truck moving uphill:* Chemical energy of fuel \rightarrow$thermal energy$\rightarrow mechanical kinetic energy and gravitational potential energy.\n * (ii) *Unwinding of a watch spring:* Elastic potential energy \rightarrow rotational kinetic energy of gears and hands.\n * (iii) *Photosynthesis in green leaves:* Solar light energy \rightarrow chemical bond energy stored in glucose.\n * (iv) *Water flowing from a dam:* Gravitational potential energy \rightarrow kinetic energy of flowing water.\n * (v) *Burning of a matchstick:* Chemical energy \rightarrow thermal energy and light energy.\n * (vi) *Explosion of a firecracker:* Chemical energy \rightarrow thermal energy, sound energy, light energy, and kinetic energy of fragments.\n * (vii) *Speaking into a microphone:* Acoustic sound energy \rightarrow electrical signal energy.\n * (viii) *Glowing electric bulb:* Electrical energy \rightarrow light energy and thermal radiation.\n * (ix) *Solar panel:* Radiant light energy \rightarrow electrical potential energy.\n * **Exercise 5 (Elevator vs. Stairs Work Comparison):**\n * *Parameters:* Student mass m = 50\,\text{kg}$; building height$h = 72.5\,\text{m}$;$g = 10\,\text{m}\,\text{s}^{-2}.\n * (i) *Elevator Gain in Potential Energy:* \Delta U = mgh = 50\,\text{kg} \times 10\,\text{m}\,\text{s}^{-2} \times 72.5\,\text{m} = 36250\,\text{J}.\n * (ii) *Staircase Gain in Potential Energy:* Elevation reached is identical (h = 72.5\,\text{m}$), so$\Delta U = mgh = 36250\,\text{J}.\n * (iii) *Conclusion:* Gravitational potential energy depends exclusively on the initial and final vertical coordinates, and is completely independent of the geometric path traversed.\n * **Exercise 6 (Crane Lifting to 10th vs. 20th Floor):**\n * Let height of 10 floors be H$, and 20 floors be$2H.\n * Energy required to 10th floor: W_1 = mgH$. Energy to 20th floor:$W_2 = mg(2H) = 2W_1 (**double the energy is required**).\n * Power to 10th floor in time t$:$P_1 = \frac{W_1}{t} = \frac{mgH}{t}.\n * Power to 20th floor in time 2t$:$P_2 = \frac{W_2}{2t} = \frac{2mgH}{2t} = \frac{mgH}{t} = P_1 (**power requirement remains unchanged**).\n * **Exercise 7 (Flagpole Work and Power):**\n * *Energy Factors:* Mass of flag and rope (m$), height of flagpole ($h$), gravitational acceleration ($g), and pulley friction.\n * *Raising Speed and Work:* Raising slowly or quickly does not alter total work done, because W = mgh is independent of time.\n * *Doubled Speed and Power:* Raising at twice the speed halves the elapsed time (t' = \frac{t}{2}$). Because$P = \frac{W}{t}$, halving time doubles the power required ($P' = 2P).\n * **Exercise 8 (Fuel Consumption Ratio for Scooter):**\n * *Day 1:* Combined mass m_1 = m_{\text{scooter}} + m_{\text{man}} = 100\,\text{kg} + 60\,\text{kg} = 160\,\text{kg}.\n * Kinetic energy gained: K_1 = \frac{1}{2}m_1 v^2 = \frac{1}{2}(160)v^2 = 80v^2.\n * *Day 2:* Combined mass m_2 = 100\,\text{kg} + 60\,\text{kg} + 40\,\text{kg} = 200\,\text{kg}.\n * Kinetic energy gained: K_2 = \frac{1}{2}m_2 v^2 = \frac{1}{2}(200)v^2 = 100v^2.\n * *Fuel Ratio:* Fuel consumed is directly proportional to required kinetic energy transfer:\n      \frac{\text{Fuel}1}{\text{Fuel}_2} = \frac{K_1}{K_2} = \frac{80v^2}{100v^2} = \frac{80}{100} = \frac{4}{5} = 0.8\n * **Exercise 9 (Balanced Seesaw with Child and Adult):**\n * *Parameters:* Adult mass m_A = 2m_C$, where$m_C is the child's mass.\n * *Balance Condition:* m_C \times d_C = m_A \times d_A \implies m_C \times d_C = (2m_C) \times d_A \implies d_C = 2d_A.\n * *Configuration:* The adult must sit at half the distance from the fulcrum compared to the child (e.g., if the child sits at 2\,\text{m}$from the fulcrum, the adult must sit at$1\,\text{m} on the opposite side).\n * **Exercise 10 (Ball Thrown Up with Air Resistance):**\n * *Parameters:* Mass m = 2\,\text{kg}$; initial speed$v_0 = 20\,\text{m}\,\text{s}^{-1}$; peak height$h = 19.4\,\text{m}$;$g = 10\,\text{m}\,\text{s}^{-2}.\n * (i) *Sign of Gravity Work:* Negative during upward ascent (gravity opposes upward displacement); positive during downward descent (gravity aligns with downward displacement).\n * (ii) *Work Done by Air Resistance:*\n * Initial kinetic energy: K_i = \frac{1}{2}(2\,\text{kg})(20\,\text{m}\,\text{s}^{-1})^2 = 400\,\text{J}.\n * Gravitational potential energy at apex: U_f = mgh = (2\,\text{kg})(10\,\text{m}\,\text{s}^{-2})(19.4\,\text{m}) = 388\,\text{J}.\n * Kinetic energy at apex: K_f = 0.\n * By work-energy theorem (W{\text{gravity}} + W_{\text{air}} = \Delta K):\n        -388\,\text{J} + W_{\text{air}} = 0 - 400\,\text{J}\n        W_{\text{air}} = -400\,\text{J} + 388\,\text{J} = -12\,\text{J}\n * Air resistance performed -12\,\text{J} of work on the ball.\n * **Exercise 11 (Block Driven by Variable Force):**\n * *Parameters:* Block mass m = 10.0\,\text{kg}$; initial kinetic energy at$0\,\text{m}$is$K_0 = 180\,\text{J}.\n * *Speed at 0\,\text{m}:*\n      K_0 = \frac{1}{2}mv_0^2 \implies 180 = \frac{1}{2}(10.0)v_0^2 \implies 5.0v_0^2 = 180 \implies v_0^2 = 36 \implies v_0 = 6\,\text{m}\,\text{s}^{-1}\n * *Speed at 4\,\text{m}:*\n * From the force-displacement graph, the triangular area from 0\,\text{m}$to$4\,\text{m}$with base$4\,\text{m}$and peak force$50\,\text{N} represents work done:\n        W = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\,\text{m} \times 50\,\text{N} = 100\,\text{J}\n * Final kinetic energy at 4\,\text{m}$:$K_4 = K_0 + W = 180\,\text{J} + 100\,\text{J} = 280\,\text{J}.\n * Final speed: \frac{1}{2}(10.0)v_4^2 = 280 \implies 5.0v_4^2 = 280 \implies v_4^2 = 56 \implies v_4 = \sqrt{56} \approx 7.48\,\text{m}\,\text{s}^{-1}.\n * *Negative Acceleration:* No. Because the applied force remains positive (F \ge 0$) across the entire range, acceleration ($a = \frac{F}{m}) is always non-negative. The block continually speeds up, though its rate of acceleration decreases after the force peak.\n * **Exercise 12 (Throwing Height on Lunar Surface):**\n * Earth peak height: h_E = \frac{v^2}{2g_E} = 8\,\text{m}.\n * Moon gravity: g_M = \frac{1}{6}g_E.\n * Lunar peak height with identical launch speed v:\n      h_M = \frac{v^2}{2g_M} = \frac{v^2}{2\left(\frac{1}{6}g_E\right)} = 6 \times \left(\frac{v^2}{2g_E}\right) = 6 \times 8\,\text{m} = 48\,\text{m}\n * **Exercise 13 (Car Braking Graph Analysis):**\n * *Parameters:* Car mass m = 1000\,\text{kg}.\n * (i) *Motion between A and B:* The car moves at a uniform, constant speed of 35\,\text{m}\,\text{s}^{-1}$for$1\,\text{second} (driver reaction delay; zero acceleration).\n * (ii) *Kinetic Energy at A:*\n      K_A = \frac{1}{2}mv^2 = \frac{1}{2}(1000\,\text{kg})(35\,\text{m}\,\text{s}^{-1})^2 = 500 \times 1225 = 612500\,\text{J}\n * (iii) *Work Done by Brakes between B and C:*\n * At C, the car is fully stopped (K_C = 0).\n * W_{\text{brakes}} = \Delta K = K_C - K_B = 0 - 612500\,\text{J} = -612500\,\text{J}.\n * (iv) *Energy Transformation:* The initial mechanical kinetic energy transforms into thermal energy in the brake pads and drums, sound waves, and road-tire frictional heat.\n * **Exercise 14 (Ball on Frictionless Track from Potential Energy Graph):**\n * *Parameters:* Mass m = 0.5\,\text{kg}$. At point O,$v_O = 0\,\text{m}\,\text{s}^{-1}$and$U_O = 30\,\text{J}.\n * Total mechanical energy: E = K_O + U_O = 0 + 30\,\text{J} = 30\,\text{J}.\n * *At Point P (U_P = 20\,\text{J}):*\n * K_P = E - U_P = 30\,\text{J} - 20\,\text{J} = 10\,\text{J}.\n * \frac{1}{2}(0.5\,\text{kg})v_P^2 = 10 \implies 0.25v_P^2 = 10 \implies v_P^2 = 40 \implies v_P = \sqrt{40} \approx 6.32\,\text{m}\,\text{s}^{-1}.\n * *At Point Q (U_Q = 0\,\text{J}):*\n * K_Q = E - U_Q = 30\,\text{J} - 0\,\text{J} = 30\,\text{J}.\n * 0.25v_Q^2 = 30 \implies v_Q^2 = 120 \implies v_Q = \sqrt{120} \approx 10.95\,\text{m}\,\text{s}^{-1}.\n * *At Point R (U_R = 10\,\text{J}):*\n * K_R = E - U_R = 30\,\text{J} - 10\,\text{J} = 20\,\text{J}.\n * 0.25v_R^2 = 20 \implies v_R^2 = 80 \implies v_R = \sqrt{80} \approx 8.94\,\text{m}\,\text{s}^{-1}.\n * **Exercise 15 (Falling Coconut Sand Impact):**\n * *Parameters:* Coconut mass m = 1.5\,\text{kg}$; tree height$h = 10\,\text{m}$; resistive sand force$F = 3000\,\text{N}$;$g = 10\,\text{m}\,\text{s}^{-2}.\n * (i) *Velocity Prior to Impact:*\n      v = \sqrt{2gh} = \sqrt{2 \times 10\,\text{m}\,\text{s}^{-2} \times 10\,\text{m}} = \sqrt{200} \approx 14.14\,\text{m}\,\text{s}^{-1}\n * (ii) *Depth of Sand Depression (x):*\n * Total mechanical energy upon reaching sand surface: E = mgh = 1.5\,\text{kg} \times 10\,\text{m}\,\text{s}^{-2} \times 10\,\text{m} = 150\,\text{J}.\n * Work done by sand resistance stops the coconut over penetration depth x:\n        F \times x = E\n        3000\,\text{N} \times x = 150\,\text{J}\n        x = \frac{150}{3000} = 0.05\,\text{m} = 5\,\text{cm}\n * The coconut penetrates 5\,\text{cm}$$ into the sand.