Comprehensive Physics Study Guide: Motion, Forces, Work, Energy, and Machines

Fundamental Physics Concepts of Motion

• Definition of Motion and Rest

  • Motion and rest are relative terms that depend entirely on the chosen fixed reference point (the observer).

  • Rest: An object is said to be at rest with respect to a reference point if its position does not change with time.

  • Motion: An object is said to be in motion with respect to a reference point if its position changes continuously with time.

• Types of Motion

  • Translational Motion: Motion in which all points of a moving body move uniformly in the same line or direction.

  • Rotational Motion: Motion where an object spins or turns around a central fixed axis.

  • Oscillatory Motion: Repetitive back-and-forth or to-and-fro motion about a central mean position.

  • Linear Motion (Motion in a Straight Line): Motion of an object along a straight path. Position on a coordinate plane is defined relative to an origin reference point $(0, 0)$, such as specifying coordinates $(3, 4)$ or $(6, 3)$.

• Scalar and Vector Quantities

  • Scalar Quantity: A physical quantity that possesses magnitude only and has no associated direction. Examples include distance and speed.

  • Vector Quantity: A physical quantity that possesses both magnitude and direction. Examples include displacement, velocity, acceleration, and force.

• Distance vs. Displacement

  • Distance:

  • Represents the total length of the actual path traversed by an object during its motion.

  • It is a scalar quantity.

  • Magnitude is always greater than or equal to displacement ($\text{Distance} \ge \text{Displacement}$).

  • Displacement:

  • Represents the shortest straight-line distance measured from the initial position to the final position of an object.

  • It is a vector quantity with a specified direction pointing from initial to final position.

  • Magnitude is always less than or equal to distance ($\text{Displacement} \le \text{Distance}$).

  • Distance and displacement are not inherently equal unless an object moves strictly along a straight line without reversing direction.

• Rates of Motion: Speed and Velocity

  • Average Speed:

  • Total path length divided by total elapsed time: $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$.

  • Scalar quantity with no direction.

  • SI Unit: $m/s$ or $m\,s^{-1}$.

  • Average Velocity:

  • Total change in position divided by total elapsed time: $\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$.

  • Vector quantity whose direction matches that of the overall displacement.

  • SI Unit: $m/s$ or $m\,s^{-1}$.

• Uniform and Non-Uniform Motion

  • Uniform Motion: An object covers equal distances in equal intervals of time. The speed remains strictly constant ($v = \text{constant}$). Example: Moving $40\,km$ every $1\,h$ ($40\,km/h$ for $1\,h \rightarrow 40\,km$, $2\,h \rightarrow 80\,km$, $3\,h \rightarrow 120\,km$).

  • Non-Uniform Motion: An object covers unequal distances in equal intervals of time. The speed changes continuously over time.

• Acceleration

  • Definition: The rate of change of velocity over time.

  • Formula: $a = \frac{v - u}{t}$, where $v$ is final velocity, $u$ is initial velocity, and $t$ is time.

  • Vector quantity.

  • SI Unit: $m/s^2$ or $m\,s^{-2}$.

  • Speeding Up (Acceleration): Velocity vector and acceleration vector point in the same direction.

  • Slowing Down (Deceleration or Retardation): Velocity vector and acceleration vector point in opposite directions.

• Graphical Analysis of Motion

  • Position-Time ($x-t$) or Distance-Time Graphs:

  • Horizontal flat line: Object at rest ($v = 0$).

  • Straight inclined line: Uniform speed ($v = \text{constant}$).

  • Curved line: Non-uniform speed / accelerated motion.

  • Slope of $x-t$ graph gives speed or velocity ($\text{Slope} = \frac{\Delta x}{\Delta t} = v$).

  • Area under $x-t$ graph carries no physical meaning ($x \times t$).

  • Velocity-Time ($v-t$) Graphs:

  • Horizontal flat line: Constant velocity ($a = 0$).

  • Straight inclined line going upward: Uniform acceleration ($a > 0$).

  • Straight line going downward: Uniform retardation/deceleration ($a < 0$).

  • Slope of $v-t$ graph gives acceleration ($\text{Slope} = \frac{\Delta v}{\Delta t} = a$).

  • Area under $v-t$ graph gives total displacement ($\text{Area} = v \times t = s$).

Kinematic Equations and Derivations

• Variables of Motion

  • Initial velocity: $u$

  • Final velocity: $v$

  • Uniform acceleration: $a$

  • Time duration: $t$

  • Displacement/Distance: $s$

  • In kinematic problems involving constant acceleration, three out of these five variables are given, and the remaining two can be solved using the three equations of motion.

• First Equation of Motion ($v = u + a\,t$)

  • Algebraic Method:

  • By definition of acceleration: $a = \frac{v - u}{t}$

  • Multiplying both sides by $t$: $a\,t = v - u$

  • Rearranging gives: $v = u + a\,t$

  • Graphical Method:

  • On a $v-t$ graph starting at velocity $u$ at $t = 0$ and reaching velocity $v$ at time $t$:

  • $\text{Slope} = a = \frac{v - u}{t}$

  • $a\,t = v - u \implies v = u + a\,t$

• Second Equation of Motion ($s = u\,t + \frac{1}{2}\,a\,t^2$)

  • Algebraic Method:

  • Average velocity for uniform acceleration: $v_{\text{avg}} = \frac{u + v}{2}$

  • Displacement $s = v_{\text{avg}} \times t = \left(\frac{u + v}{2}\right) \times t$

  • Substitute $v = u + a\,t$ into displacement equation: $s = \left(\frac{u + u + a\,t}{2}\right) \times t = \left(\frac{2\,u + a\,t}{2}\right) \times t$

  • Simplifying gives: $s = u\,t + \frac{1}{2}\,a\,t^2$

  • Graphical Method:

  • Displacement $s$ equals total area under the $v-t$ graph between $0$ and $t$:

  • $\text{Area} = \text{Area of Rectangle} + \text{Area of Triangle}$

  • $s = (u \times t) + \frac{1}{2} \times t \times (v - u)$

  • Substitute $(v - u) = a\,t$: $s = u\,t + \frac{1}{2}\,a\,t^2$

• Third Equation of Motion ($2\,a\,s = v^2 - u^2$)

  • Algebraic Method:

  • Displacement: $s = \left(\frac{u + v}{2}\right) \times t$

  • From first equation, $t = \frac{v - u}{a}$

  • Substitute $t$ into displacement equation: $s = \left(\frac{v + u}{2}\right) \times \left(\frac{v - u}{a}\right) = \frac{(v + u)(v - u)}{2\,a}$

  • Cross-multiplying gives: $2\,a\,s = v^2 - u^2$

  • Graphical Method:

  • Displacement $s$ equals area of trapezium formed under $v-t$ graph:

  • $\text{Area of Trapezium} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height}$

  • $s = \frac{1}{2} \times (v + u) \times t$

  • Substitute $t = \frac{v - u}{a}$: $s = \frac{(v + u)(v - u)}{2\,a}$

  • Rearranging yields: $2\,a\,s = v^2 - u^2$

• Uniform Circular Motion

  • Definition: Motion of an object moving along a circular path at a constant speed.

  • Time Period ($T$): Time taken to complete one full revolution.

  • Path length for one revolution equals circumference: $2\,\pi\,R$

  • Speed equation: $v = \frac{2\,\pi\,R}{T}$

  • Why Uniform Circular Motion is Accelerated:

  • Acceleration is defined as any change in velocity over time.

  • Velocity changes if magnitude (speed) changes, direction changes, or both change.

  • In uniform circular motion, speed is constant, but the direction of motion changes continuously at every point along the circular path.

  • Continuous change in direction causes continuous change in velocity, producing a non-zero acceleration known as centripetal acceleration directed towards the circle's center.

Dynamics and Forces

• Concept of Force

  • A force is a push or pull exerted on an object that results from its interaction with another object.

  • Vector quantity requiring both magnitude and direction.

  • Measured using a Spring Balance (used to measure force and weight).

  • SI Unit: Newton ($N$).

  • Effects of Force:

  • Move an object from rest.

  • Change the speed of a moving object.

  • Change the direction of motion of an object.

  • Alter the shape or size of an object.

• Balanced vs. Unbalanced Forces

  • Balanced Forces:

  • Two or more forces equal in magnitude and opposite in direction acting on an object.

  • Net external force $F_{\text{net}} = 0$

  • Acceleration $a = 0$

  • Effect: Does not change state of rest ($u = 0, v = 0$) or uniform motion ($u = 5\,m/s, v = 5\,m/s$).

  • Unbalanced Forces:

  • Forces acting on an object where combined net vector force is non-zero.

  • Net external force $F_{\text{net}} \neq 0$

  • Acceleration $a \neq 0$

  • Effect: Starts motion, accelerates, decelerates, or changes direction.

• Net Force Calculations

  • Forces acting in the same direction: $F_{\text{net}} = F_1 + F_2$ (directed along the forces).

  • Forces acting in opposite directions: $F_{\text{net}} = F_{\text{larger}} - F_{\text{smaller}}$ (directed toward the larger force).

• Force of Friction

  • Contact force acting between contact surfaces that always opposes the relative motion of objects.

  • Characterized by slowing down moving bodies and converting kinetic energy to thermal energy.

  • To maintain motion at constant velocity ($v = \text{constant}$), an applied forward force must continuously balance the frictional force ($F_{\text{applied}} = F_{\text{friction}}$).

Newton's Laws of Motion and System Mechanics

• Linear Momentum ($p$)

  • Product of mass and velocity of a body: $p = m\,v$

  • Vector quantity with direction matching velocity.

  • SI Unit: kilogram metre per second ($kg\,m/s$ or $kg\,m\,s^{-1}$).

  • Photons travel at light speed ($3 \times 10^8\,m/s$) with zero rest mass ($m = 0$).

• Newton's First Law of Motion (Law of Inertia)

  • Statement: A body at rest remains at rest, and a body in motion continues in uniform motion along a straight line, unless acted upon by an external unbalanced force.

  • Mathematical statement: If $F = 0$, then $a = 0$, meaning velocity $v$ remains constant.

  • Inertia: Inherent property of matter to resist any change in its state of rest or uniform motion.

• Newton's Second Law of Motion

  • Statement: The rate of change of momentum of an object is directly proportional to the net applied unbalanced force and takes place in the direction of the force.

  • Mathematical formulation: $F \propto \frac{\Delta p}{t} \implies F = k \frac{\Delta p}{t}$ (taking constant $k = 1$).

  • Derivation of $F = m\,a$:

  • Initial momentum $p_i = m\,u$

  • Final momentum $p_f = m\,v$

  • Change in momentum $\Delta p = p_f - p_i = m\,v - m\,u$

  • Force $F = \frac{m\,v - m\,u}{t} = m\left(\frac{v - u}{t}\right)$

  • Since $a = \frac{v - u}{t}$, it follows that $F = m\,a$

  • Impact Time Relationship:

  • Force $F = \frac{\Delta p}{t}$. Increasing impact duration ($t \uparrow$) decreases the force experienced ($F \downarrow$).

• Newton's Third Law of Motion

  • Statement: For every action force, there is an equal and opposite reaction force.

  • Action and reaction forces act on two different bodies simultaneously.

• Forces on Connected Systems (Multi-Body Mechanics)

  • For two connected objects of mass $m_1$ (Box 1) and $m_2$ (Box 2) joined by a string:

  • External force $F$ acts on Box 1 to the right, internal string tension $T$ acts between them.

  • Force breakdown on Box 1: Applied force $F$ rightward, tension $T$ leftward.

  • Force breakdown on Box 2: Tension $T$ rightward.

  • System behaves as a single object of total mass $(m_1 + m_2)$.

  • System Acceleration Formula: $a = \frac{F_{\text{ext}}}{m_1 + m_2}$

Work, Energy, and Power

• Work ($W$)

  • Work is defined as the product of force and displacement produced in the direction of force.

  • Formula: $W = F \times s$

  • SI Unit: Joule ($J$). $1\,\text{Joule} = 1\,\text{Newton} \times 1\,\text{metre}$ ($1\,J = 1\,N\cdot m$).

  • Conditions for Work Types:

  • Positive Work: Force and displacement act in the same direction ($W > 0$). Example: Lifting a weight upward (applied force is upward).

  • Negative Work: Force and displacement act in opposite directions ($W < 0$). Example: Lowering a weight or applying brakes to a car.

  • Zero Work: Work $W = 0$ when force $F = 0$, displacement $s = 0$ (e.g., pushing a fixed wall), or force is perpendicular to displacement ($F \perp s$).

• Work-Energy Theorem

  • Work done on a body equals the net change in its total energy ($W = \Delta E$).

• Forms of Energy

  • Kinetic Energy ($K$ or $KE$):

  • Energy possessed by an object due to its state of motion.

  • Formula: $K = \frac{1}{2}\,m\,v^2$

  • Scalar quantity depending on mass $m$ and velocity $v$.

  • SI Unit: Joule ($J$).

  • Potential Energy ($U$ or $PE$):

  • Energy possessed by an object due to its position, height, or elastic deformation.

  • Gravitational Potential Energy Formula: $PE = m\,g\,h$ (where $g = 9.8\,m/s^2$ or $10\,m/s^2$).

  • Elastic Potential Energy: Energy stored in deformed bodies (e.g., compressed spring or stretched band).

  • Mechanical Energy ($E$):

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

  • Conservation of Mechanical Energy:

  • In an isolated system subject only to conservative forces (like gravity), total mechanical energy remains constant ($E = U + K = \text{constant}$).

  • Free Fall Example ($m = 5\,kg$, height $h = 10\,m$, $g = 10\,m/s^2$):

    • At maximum height ($h = 10\,m$, initial velocity $u = 0$):

    • $U = m\,g\,h = 5 \times 10 \times 10 = 500\,J$

    • $K = 0\,J$

    • $E = 500 + 0 = 500\,J$

    • Work done to raise mass to height $h$: $W = F \times s = (m\,g) \times h = 50 \times 10 = 500\,J$

    • At bottom ground position ($h = 0\,m$):

    • Potential energy drops to $U = 0\,J$

    • Kinetic energy increases to $K = 500\,J$

    • Total mechanical energy $E = U + K = 500\,J$

• Power ($P$)

  • Rate at which work is performed or energy is converted over time.

  • Formula: $P = \frac{W}{t}$

  • SI Unit: Watt ($W$).

  • $1\,\text{Watt} = 1\,\text{Joule/second}$ ($1\,W = 1\,J/s$).

Simple Machines and Mechanical Advantage

• Function of Simple Machines

  • Simple devices that make work easier by altering the magnitude, direction, or point of application of an applied force.

  • Effort ($E$): Applied force input to the machine.

  • Load ($L$): Resistance or weight force overcome by the machine.

  • Mechanical Advantage ($MA$): Ratio of load force overcome to effort force applied:

  • $MA = \frac{\text{Load}}{\text{Effort}}$

• Pulley

  • Wheel with a groove along its rim holding a cable or rope.

  • Fixed Pulley: Changes the direction of effort force. Mechanical advantage $MA = 1$ (meaning $\text{Effort} = \text{Load}$).

• Inclined Plane

  • Sloped rigid surface used to lift heavy loads vertically with reduced effort force.

  • Sacrifices distance for force reduction: longer path length requires proportionally smaller effort force to elevate a load against gravity.

• Lever and Torque Balancing

  • Rigid bar rotating around a fixed pivot point called a Fulcrum.

  • Components: Fulcrum, Load Arm (distance from load to fulcrum), Effort Arm (distance from effort to fulcrum).

  • Torque Balancing / Law of Moments: For a lever system in rotational equilibrium, total clockwise moment equals total counter-clockwise moment:

  • $m\,g\,d_1 = M\,g\,d_2$

  • Example: Seesaw balancing a child ($m\,g$ at distance $d_1$) against an adult ($M\,g = 1900\,N$ from $190\,kg$ at distance $d_2$).

Exhaustive Solved Problems and Conceptual Exercises

• Motion Solved Problems

  • Question: Which of the following is a vector quantity?

  • Options: A) Distance, B) Speed, C) Displacement, D) Time

  • Answer: C) Displacement

  • Question: A car travels $150\,m$ in $10\,s$. Its average speed is:

  • Options: A) $10\,m/s$, B) $15\,m/s$, C) $20\,m/s$, D) $25\,m/s$

  • Calculation: $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{150\,m}{10\,s} = 15\,m/s$

  • Answer: B) $15\,m/s$

  • Question: A body moves with initial velocity $10\,m/s$ and reaches $30\,m/s$ in $5\,s$. Acceleration is:

  • Options: A) $2\,m/s^2$, B) $4\,m/s^2$, C) $5\,m/s^2$, D) $8\,m/s^2$

  • Calculation: $a = \frac{v - u}{t} = \frac{30 - 10}{5} = \frac{20}{5} = 4\,m/s^2$

  • Answer: B) $4\,m/s^2$

  • Question: The slope of a velocity-time graph gives:

  • Options: A) Speed, B) Displacement, C) Acceleration, D) Distance

  • Answer: C) Acceleration

  • Question: A train starts from rest and reaches a velocity of $20\,m/s$ in $10\,s$. What is its acceleration?

  • Options: A) $1\,m/s^2$, B) $2\,m/s^2$, C) $10\,m/s^2$, D) $20\,m/s^2$

  • Calculation: $u = 0\,m/s$, $v = 20\,m/s$, $t = 10\,s$. $a = \frac{20 - 0}{10} = 2\,m/s^2$

  • Answer: B) $2\,m/s^2$

  • Problem: A car starts from rest and its velocity reaches $24\,m/s$ in $6\,s$. Find average acceleration and distance travelled.

  • Given: $u = 0\,m/s$, $v = 24\,m/s$, $t = 6\,s$

  • Step 1: Average acceleration $a = \frac{v - u}{t} = \frac{24 - 0}{6} = 4\,m/s^2$

  • Step 2: Distance $s = u\,t + \frac{1}{2}\,a\,t^2 = (0 \times 6) + \frac{1}{2}(4)(6)^2 = 2 \times 36 = 72\,m$

  • Alternative Method: $2\,a\,s = v^2 - u^2 \implies 2(4)s = 24^2 - 0^2 \implies 8\,s = 576 \implies s = 72\,m$

  • Answers: Acceleration = $4\,m/s^2$, Distance = $72\,m$

  • Problem: A motorbike moving with initial velocity $28\,m/s$ and constant acceleration stops after travelling $98\,m$. Find acceleration and time taken to stop.

  • Given: $u = 28\,m/s$, $v = 0\,m/s$, $s = 98\,m$

  • Acceleration calculation: $v^2 = u^2 + 2\,a\,s \implies 0^2 = 28^2 + 2(a)(98) \implies 0 = 784 + 196\,a \implies a = -\frac{784}{196} = -4\,m/s^2$

  • Time calculation: $v = u + a\,t \implies 0 = 28 + (-4)t \implies 4\,t = 28 \implies t = 7\,s$

  • Answers: Acceleration = $-4\,m/s^2$ (deceleration of $4\,m/s^2$), Time = $7\,s$

  • Problem: A truck driver driving at $54\,km/h$ notices a speed limit sign of $40\,km/h$ and slows down to $36\,km/h$ in $36\,s$. What distance was travelled during this time?

  • Given: $u = 54\,km/h = 54 \times \frac{5}{18} = 15\,m/s$, $v = 36\,km/h = 36 \times \frac{5}{18} = 10\,m/s$, $t = 36\,s$

  • Step 1: Acceleration $a = \frac{v - u}{t} = \frac{10 - 15}{36} = -\frac{5}{36}\,m/s^2$

  • Step 2: Distance $2\,a\,s = v^2 - u^2 \implies 2\left(-\frac{5}{36}\right)s = 10^2 - 15^2 \implies -\frac{10}{36}\,s = 100 - 225 = -125$

  • $s = \frac{125 \times 36}{10} = 450\,m$

  • Answer: Distance travelled = $450\,m$

  • Problem: A car starts from rest and accelerates uniformly to $20\,m/s$ in $5\,s$. It then travels at $20\,m/s$ for $10\,s$ and finally applies brakes with uniform acceleration to stop in $6\,s$. Find total distance travelled.

  • Case 1 (Acceleration): $u_1 = 0\,m/s$, $v_1 = 20\,m/s$, $t_1 = 5\,s$

    • $a_1 = \frac{20 - 0}{5} = 4\,m/s^2$

    • $s_1 = \frac{v_1^2 - u_1^2}{2\,a_1} = \frac{20^2 - 0}{2(4)} = \frac{400}{8} = 50\,m$

  • Case 2 (Constant Speed): $v = 20\,m/s$, $t_2 = 10\,s$

    • $s_2 = v \times t_2 = 20 \times 10 = 200\,m$

  • Case 3 (Deceleration): $u_3 = 20\,m/s$, $v_3 = 0\,m/s$, $t_3 = 6\,s$

    • $a_3 = \frac{0 - 20}{6} = -\frac{10}{3}\,m/s^2$

    • $2\,a_3\,s_3 = v_3^2 - u_3^2 \implies 2\left(-\frac{10}{3}\right)s_3 = 0^2 - 20^2 \implies -\frac{20}{3}\,s_3 = -400 \implies s_3 = 60\,m$

  • Total Distance: $s_{\text{total}} = s_1 + s_2 + s_3 = 50\,m + 200\,m + 60\,m = 310\,m$

  • Answer: Total distance = $310\,m$

  • Problem: A bus is travelling at $36\,km/h$ when the driver sees an obstacle $30\,m$ ahead. Driver reaction time is $0.5\,s$. Once brakes are applied, retardation is $2.5\,m/s^2$. Will the bus stop before reaching the obstacle?

  • Given: $u = 36\,km/h = 10\,m/s$, $d_{\text{obstacle}} = 30\,m$, $t_{\text{reaction}} = 0.5\,s$, $a = -2.5\,m/s^2$

  • Step 1: Distance in reaction time $s_1 = u \times t_{\text{reaction}} = 10 \times 0.5 = 5\,m$

  • Step 2: Braking distance $s_2$ when stopping ($v = 0\,m/s$):

    • $v^2 = u^2 + 2\,a\,s_2 \implies 0^2 = 10^2 + 2(-2.5)s_2 \implies 0 = 100 - 5\,s_2 \implies s_2 = 20\,m$

  • Step 3: Total stopping distance $s_{\text{total}} = s_1 + s_2 = 5\,m + 20\,m = 25\,m$

  • Comparison: Available distance is $30\,m$. Since $25\,m < 30\,m$, the bus stops safely $30\,m - 25\,m = 5\,m$ before reaching the obstacle.

  • Answer: Yes, the bus stops $5\,m$ before the obstacle.

• Forces Solved Problems

  • Question: A force can change the motion of an object by changing its:

  • Options: A) Shape only, B) Speed or direction, C) Mass only, D) Volume only

  • Answer: B) Speed or direction

  • Question: When two or more forces act on an object in the same direction, the net force is:

  • Options: A) Zero, B) The difference between the forces, C) The sum of the forces, D) Always equal to the smaller force

  • Answer: C) The sum of the forces

  • Question: A force of $10\,N$ acts towards the right and $6\,N$ acts towards the left. Net force is:

  • Options: A) $16\,N$ right, B) $16\,N$ left, C) $4\,N$ right, D) $4\,N$ left

  • Calculation: $F_{\text{net}} = F_{\text{right}} - F_{\text{left}} = 10 - 6 = 4\,N$ towards right

  • Answer: C) $4\,N$ towards right

  • Question: A car travels $120\,km$ in $3\,h$. Its average speed is:

  • Options: A) $30\,km/h$, B) $40\,km/h$, C) $60\,km/h$, D) $360\,km/h$

  • Calculation: $\text{Average Speed} = \frac{120\,km}{3\,h} = 40\,km/h$

  • Answer: B) $40\,km/h$

  • Assertion / Reason Question:

  • Assertion (A): A moving object can continue moving with constant velocity even when no net force acts on it.

  • Reason (R): An object tends to maintain its state of motion due to inertia.

  • Answer: Both A and R are true, and R is the correct explanation of A.

  • Question: Using horizontal force $F$, a table moves across the floor at constant velocity. What is the frictional force?

  • Answer: At constant velocity, net force $F_{\text{net}} = 0$. Therefore, frictional force equals applied force ($F_{\text{friction}} = F$).

  • Question: Block P experiences opposing forces $4\,N$ and $5\,N$. Block Q moves with constant velocity. Which experiences a net force?

  • Calculation: For P, $F_{\text{net}} = 5\,N - 4\,N = 1\,N$. For Q, constant velocity means $a = 0 \implies F_{\text{net}} = 0$.

  • Answer: P experiences a net force ($1\,N$) and Q does not experience a net force.

  • Question: Position-time graphs for four objects A, B, C, D: A is at rest, B moves with constant velocity, C has a curved line (changing velocity), D moves with constant velocity. A net force acts on:

  • Answer: Object C, because its non-linear graph indicates changing velocity (acceleration), requiring a net external force ($F \neq 0$).

  • Question: A sailor jumps forward from a small boat to the shore. Will the boat move? In which direction and why?

  • Answer: Yes, the boat moves backward away from the shore. By Newton's Third Law, the sailor pushing forward on the shore exerts an equal and opposite reaction force pushing the boat backward.

  • Question: Why is a landing mat or sand bed placed in a high jump event?

  • Answer: An athlete landing from a high jump possesses high momentum. A soft mat increases impact time ($t \uparrow$). By Newton's Second Law ($F = \frac{\Delta p}{t}$), increasing impact time reduces impact force ($F \downarrow$), preventing injury.

  • Problem: Velocity-time graph of $10\,kg$ mass shows velocity increases from $10\,m/s$ at $t = 0\,s$ to $30\,m/s$ at $t = 8\,s$. Calculate force.

  • Given: $m = 10\,kg$, $u = 10\,m/s$, $v = 30\,m/s$, $t = 8\,s$

  • Step 1: Acceleration $a = \frac{v - u}{t} = \frac{30 - 10}{8} = \frac{20}{8} = 2.5\,m/s^2$

  • Step 2: Force $F = m\,a = 10 \times 2.5 = 25\,N$

  • Answer: Force = $25\,N$

• Work, Energy, and Machines Solved Problems

  • Question: Work is done when force causes:

  • Options: A) Change in shape only, B) Displacement in direction of force, C) No movement, D) Change in temperature

  • Answer: B) Displacement in the direction of force

  • Question: SI unit of work and energy is:

  • Options: A) Watt, B) Newton, C) Joule, D) Pascal

  • Answer: C) Joule

  • Question: Energy possessed by a body due to motion is:

  • Options: A) Potential energy, B) Kinetic energy, C) Heat energy, D) Chemical energy

  • Answer: B) Kinetic energy

  • Question: A simple machine is mainly used to:

  • Options: A) Create energy, B) Make work easier by changing magnitude or direction of force, C) Reduce amount of work done, D) Increase mass

  • Answer: B) Make work easier by changing magnitude or direction of force

  • Question: Which of the following is an example of a lever?

  • Options: A) Pulley, B) Inclined plane, C) Seesaw, D) Screw

  • Answer: C) Seesaw

  • True or False Statements:

  • (i) Work is done when force is applied even if object does not move. [False]

  • (ii) Lifting a bucket vertically upward results in positive work done on bucket. [True]

  • (iii) The SI unit for both work and energy is joule ($J$). [True]

  • (iv) A motionless stretched rubber band has kinetic energy. [False]

  • (v) Energy can change from one form to another. [True]

  • Fill in the Blanks:

  • (i) Work done = Force $\times$ displacement (in direction of force).

  • (ii) $1\,\text{joule}$ of work is done when force of $1\,\text{newton}$ displaces an object by $1\,\text{metre}$ in direction of force.

  • (iii) Kinetic energy expression for mass $m$ and velocity $v$ is $\frac{1}{2}\,m\,v^2$.

  • (iv) Potential energy of mass $m$ at height $h$ is $U =$ $m\,g\,h$.

  • (v) Power is defined as the rate at which work is done.

  • Problem: Seesaw balancing: Child sits on one side, adult weighing twice the child ($W_A = 2\,W_C$) on the other. For balance, find distance ratio.

  • Torque balance equation: $W_C \times d_C = W_A \times d_A$

  • Substitute $W_A = 2\,W_C$: $W_C \times d_C = 2\,W_C \times d_A \implies d_C = 2\,d_A$

  • Answer: Child must sit at double the distance from fulcrum compared to adult ($d_C = 2\,d_A$).

  • Problem: A ball of mass $2\,kg$ is thrown up with velocity $20\,m/s$.

  • (i) Sign of work done by gravity: Upward motion = Negative; Downward motion = Positive.

  • (ii) If ball reaches maximum height $h = 19.4\,m$, find work done by air resistance ($g = 10\,m/s^2$).

    • Initial kinetic energy: $K_i = \frac{1}{2}\,m\,u^2 = \frac{1}{2}(2)(20)^2 = 400\,J$

    • Gain in potential energy: $U = m\,g\,h = (2)(10)(19.4) = 388\,J$

    • Work done by air resistance: $W_{\text{air}} = PE - KE = 388\,J - 400\,J = -12\,J$

    • Answer: Work done by air resistance = $-12\,J$

  • Problem: A $1000\,kg$ car moves at constant speed $35\,m/s$ between A and B, then brakes to halt between B and C.

  • (i) Motion between A and B: Uniform motion with constant speed ($35\,m/s$).

  • (ii) Kinetic energy at A:

    • $K = \frac{1}{2}\,m\,v^2 = \frac{1}{2}(1000)(35)^2 = 500 \times 1225 = 612500\,J = 6.125 \times 10^5\,J$

  • (iii) Work done by brakes between B and C:

    • Car comes to complete stop ($v = 0\,m/s$).

    • $W = \Delta K = K_f - K_i = 0 - 612500\,J = -612500\,J$

  • (iv) Transformation of kinetic energy: Converts into heat energy and sound energy due to braking friction.