Work, Energy, and Simple Machines (Class 9): Notes और MCQ Practice

आसान भाषा में

Science में Work तभी माना जाता है जब किसी object पर Force लगे और उसमें displacement भी हो। किसी काम को करने की capacity को Energy कहते हैं। जब object motion में होता है तो Kinetic Energy होती है, और shape या position बदलने पर Potential Energy store होती है!

Key Points & Formulas

• Work Done: W = F × s (SI Unit: Joule, 1 J = 1 N × 1 m = 1 kg m² s⁻²)
• Zero Work: अगर displacement s = 0 हो (जैसे rigid wall को धक्का देना) या Force और displacement perpendicular हों (जैसे box लेकर सीधा चलना)।
• Positive vs Negative Work:
• Force और displacement same direction में हों → Positive Work (जैसे wheelchair को push करना)।
• Force और displacement opposite direction में हों → Negative Work (जैसे goalkeeper द्वारा ball को रोकना)।
• Work-Energy Theorem: Work done on an object = Change in its energy = (1 ÷ 2) × m × (v² − u²)
• Kinetic Energy (गतिज ऊर्जा): K = (1 ÷ 2) × m × v²
• Potential Energy: Stored energy जो deformation (stretched bow, spring) या relative position (magnets, charges) की वजह से होती है।

2 Memory Tricks

1. "No Move, Zero Groove": अगर displacement = 0 है, तो Physics में Work = 0 होगा, चाहे आप थक ही क्यों न जाएं!
2. "Double Speed, 4-Times Feed": Kinetic Energy formula में v² है—speed double (2×) होगी तो Kinetic energy सीधे 4 गुना (4×) बढ़ेगी।

2 Common Board-Exam Mistakes

1. Sign Mistake in Work: Retarding force या stopping force के case में displacement opposite होने पर Work Negative (−) होता है, minus sign भूलना नहीं है।
2. Perpendicular Force Confusion: जब applied force और displacement 90° (perpendicular) पर हों, तब work done zero होता है, न कि F × s!

Work, Energy, and Simple Machines

1. Chapter in Short

This chapter introduces the scientific meanings of work and energy:

• In everyday life, we use "work" loosely, but in science, work is only done when a force causes an object to undergo displacement.
• Energy is defined as the capacity to do work. Doing positive work on an object increases its energy, as described by the Work-Energy Theorem.
• The chapter examines different forms of energy, focusing closely on Mechanical Energy, which consists of:
• Kinetic Energy: energy possessed by an object due to its motion.
• Potential Energy: energy stored in an object or system due to its shape, deformation, or relative position.

2. Key Concepts Explained Simply

Scientific Definition of Work

• In science, work requires two things: a force must act on an object, and the object must move (be displaced).
• When a constant force acts along the line of motion:

$\text{Work} = \text{Force} \times \text{Displacement in the direction of force}$

• Everyday Example: Lifting a 5 kg bag of wheat by 1 m requires doing work against gravity. Lifting 3 bags to 1 m, or 1 bag to 3 m, requires 3 times as much work.

When is Work Done Zero?

1. Zero Force ($F = 0$): No force is applied.
2. Zero Displacement ($s = 0$): A force is applied, but the object does not move. For example, pushing hard against a rigid wall does zero work on the wall, even though your muscles get tired from expanding and contracting.
3. Perpendicular Force and Displacement: When the force acts at a right angle (90°) to the direction of motion. For example, a girl carrying a box walks horizontally while applying an upward force to balance its weight. The work done on the box by her upward lifting force is zero.

Positive vs. Negative Work

• Positive Work: Displacement is in the same direction as the force.

Example: Pushing a wheelchair forward; lifting a dumbbell upward.

• Negative Work: Displacement is opposite to the direction of the force.

Example: A goalkeeper applying a backward force to stop an incoming football; lowering a dumbbell slowly downward.

Work from a Force-Displacement Graph

• The work done on an object by a force equals the area under the force-displacement graph between the initial and final positions.

The Work-Energy Theorem

• Work done on an object causes a change in its energy:

$\text{Work done on an object} = \text{Change in its energy}$

• Doing positive work adds energy to an object; doing negative work takes energy away from it.
• Example: In carrom, a moving striker does positive work on a coin, increasing the coin's kinetic energy, while the coin exerts an equal and opposite force doing negative work on the striker, slowing it down.

Forms of Energy

• Energy exists in various forms: Mechanical, Thermal, Light, Sound, Electrical, Nuclear, and Chemical.
• Energy can convert from one form to another (e.g., electrical to light and heat in a bulb, chemical energy in food to mechanical energy in muscles).

Kinetic Energy ($K$)

• The energy an object possesses because of its motion. Any object at rest has zero kinetic energy.
• For an object of mass $m$ moving with velocity $v$:

$K = \frac{1}{2} m v^2$

• If the speed of an object doubles, its kinetic energy becomes 4 times greater.

Potential Energy

• Energy stored in an object or a system due to its configuration, shape, or position:
• Due to shape/deformation: A stretched elastic band of a slingshot, a bent archery bow, or a compressed spring.
• Due to relative position in a system: Two separated unlike magnetic poles, separated electric charges, or a ball raised above the Earth's surface.

3. Important Definitions, Formulas, Facts, and Units

• Work Done ($W$): $W = F \times s$
• SI Unit of Work: joule (J). $1\text{ J} = 1\text{ N} \times 1\text{ m} = 1\text{ kg m}^2\text{ s}^{-2}$.
• Definition of 1 Joule: The work done on an object when a constant force of $1\text{ N}$ displaces it by $1\text{ m}$ in the direction of the force.
• Work-Energy Theorem: $\text{Work done} = \text{Change in energy} = \frac{1}{2} m (v^2 - u^2)$.
• Kinetic Energy Formula: $K = \frac{1}{2} m v^2$.
• Scalar Nature: Both work and kinetic energy have magnitude and a sign (positive or negative for work), but no direction.
• Scientist: The unit joule is named after James Prescott Joule, who studied the relationship between mechanical and thermal energy.

4. Solved Examples

Example 1: Calculating Work Done in Stopping an Object

Question: A goalkeeper stops a football by exerting a backward force of $200\text{ N}$. Her hands move back by $15\text{ cm}$ while stopping the ball. How much work did the goalkeeper do on the ball?

• Given: Force $F = 200\text{ N}$, Displacement $s = -15\text{ cm} = -0.15\text{ m}$ (negative because it opposes the applied force direction).
• Formula: $W = F \times s$
• Calculation:

$W = 200\text{ N} \times (-0.15\text{ m}) = -30\text{ J}$

• Answer: The goalkeeper did $-30\text{ J}$ of work on the ball.

Example 2: Effect of Speed on Kinetic Energy

Question: If the velocity of a vehicle doubles, what happens to its kinetic energy?

• Given: Initial velocity $= v$, New velocity $= 2v$, Mass $= m$.
• Formula: $K = \frac{1}{2} m v^2$
• Calculation:

$K_{\text{initial}} = \frac{1}{2} m v^2$

$K_{\text{new}} = \frac{1}{2} m (2v)^2 = \frac{1}{2} m (4v^2) = 4 \times \left(\frac{1}{2} m v^2\right) = 4 K_{\text{initial}}$

• Answer: The kinetic energy increases to $4\text{ times}$ its original value.

Example 3: Finding Kinetic Energy of a Cricket Ball

Question: A cricket ball of mass $0.2\text{ kg}$ is bowled at a speed of $154.8\text{ km h}^{-1}$. Calculate its kinetic energy.

• Conversion: $v = 154.8 \times \frac{5}{18} = 43\text{ m s}^{-1}$
• Formula: $K = \frac{1}{2} m v^2$
• Calculation:

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

• Answer: The kinetic energy of the ball is $184.9\text{ J}$.

Example 4: Work-Energy Theorem on a Landing Jet

Question: A jet of mass $15000\text{ kg}$ is stopped on a carrier deck over a distance of $100\text{ m}$ by an arresting wire providing an opposing force of $367500\text{ N}$. What was the velocity of the aircraft just before the hook engaged?

• Given: $m = 15000\text{ kg}$, $s = -100\text{ m}$ (opposing force), $F = 367500\text{ N}$, final velocity $v_{\text{final}} = 0$.
• Formula: $\text{Change in Kinetic Energy} = \text{Work done}$
• Calculation:

$0 - \frac{1}{2} m v^2 = F \times (-s)$

$-\frac{1}{2} \times 15000 \times v^2 = -(367500 \times 100)$

$7500 \times v^2 = 36750000$

$v^2 = \frac{36750000}{7500} = 4900$

$v = \sqrt{4900} = 70\text{ m s}^{-1} \quad (\text{or } 252\text{ km h}^{-1})$

• Answer: The aircraft had an initial velocity of $70\text{ m s}^{-1}$.

5. Common Mistakes and Quick Revision Points

• Feeling tired does not mean work is done: Pushing a wall does not displace it ($s = 0$), so the work done on the wall is strictly $0\text{ J}$.
• Perpendicular forces do no work: Carrying a bag while walking horizontally means the vertical lifting force does zero work on the bag because the displacement has no component in the direction of the lifting force.
• Check units before calculation: Always convert speeds from $\text{km h}^{-1}$ to $\text{m s}^{-1}$ (multiply by $\frac{5}{18}$) and distances from $\text{cm}$ to $\text{m}$ before using energy and work formulas.
• Sign of work: Work is negative whenever the force opposes displacement (e.g., friction, stopping forces).
• Direction vs. Scalar: Force and displacement are vector quantities (they have direction), but work and energy are scalar quantities (they have no direction).