9th Sci. Work, Energy, and Power—Notes

9th Sci. Work, Energy, and Power—Note

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✍🏻... विज्ञान विषयाच्या माहितीसाठी खालील⬇️ लिंकला🖇️ स्पर्श करा.

   👉  माहिती विज्ञानाची 🎷

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1. Work

Definition: Work is done when a force applied on an object causes displacement in the direction of the force.

Formula: W = F. s 

or 

W = F s cos θ 

when force is at an angle θ to displacement.

Units:

SI unit: Joule (J)

1 J = 1 N . 1 m

CGS unit: Erg 

1 erg = 1 dyne. 1 cm

Relation: 

1 J = 10⁷ ergs

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🧑‍🔧 Types of Work:

➕ Positive Work (θ = 0°): Force and displacement are in the same direction.

➖ Negative Work (θ = 180°): Force and displacement are in opposite directions (e.g., work done against gravity or friction).

🅾️ Zero Work (θ = 90° or s = 0): Force is perpendicular to displacement, or no motion occurs (e.g., circular orbit of a satellite).

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2. 💪Energy

Definition: The capacity of a body to perform work. A unit is the same as work (joule).

Kinetic Energy (K.E.): Energy possessed due to motion.

K.E. = 1/2 m v²

Potential Energy (P.E.): Energy possessed by a body due to its position, shape, or configuration is called potential energy.

Or 

Energy stored due to position or state.

P.E = m g h

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⛽ Law of Conservation of Energy: Energy can neither be created nor destroyed; it can only be converted from one form to another. Total energy in the universe remains constant.

🍃 Free Fall Energy Conservation: At any point during a free fall, total energy (T.E.) remains constant: 

T.E. = P.E. + K.E. = m g h.

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👊 Power

Definition: Rate at which work is done or energy is consumed.

Formula: P = W/t

Units:

SI unit: Watt (W) 

1 W = 1 J / s

Industrial unit: Horsepower (hp)

1 hp = 746 W

Commercial Unit of Electrical Energy:

Kilowatt-hour (kWh) or "unit"

1 kWh = 1000 W × 3600 s 

∴ 1 kWh = 3.6 ×10⁶ J

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Example 

Problem: Calculate the work done to lift an object of mass 30 kg to a height of 15 m.

(g = 9.8 m/s²).

Solution:

Given: 

m = 30 kg, s = 15 m, g = 9.8 m/s²

Force applied against gravity: 

F = -m g = -30 × 9.8 = -294 N

Work done: 

W = F . s = -294 / 15 = -4410 J

(The negative sign indicates force opposes direction of motion).

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Example 

Problem: A force of 150 N is applied to an object at an angle of 60 to the horizontal. If 600 J of work is done, find the displacement. (cos 60° = 0.5)

Solution:

Given: 

F = 150 N, θ = 60*, W = 600 J

Formula: W = F. s cos θ

600 = 150 × s × 0.5

600 = 75 × s

s = 600 / 75

s = 8 m

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Example

Problem: An object having a mass of 400 g is moving with a velocity of 5 m/s. Calculate its kinetic energy.

Solution:

Given: m = 400 g = 0.4 kg, v = 5 m/s

Formula: 

K.E. = 1/2 m v²

K.E. = 1/2 × 0.4 × 5² 

= 0.2 × 25

K.E. = 5 J 

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Example 

Problem: Swaralee takes 30 s to carry a bag weighing 25 kg to a height of 6 m. How much power has she used? (g = 9.8 m/s²)

Solution:

Given: 

m = 25 kg, h = 6 m, t = 30 

F = mg

= 25 × 9.8

= 245 N

W = F . s

= 245 × 6

= 1470 J

P = W/t = 1470/30 = 49 W

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Example

Problem: 800 kg of water is stored in an overhead tank at a height of 12 m. Calculate the potential energy stored. (g = 9.8 m/s²)

Given: 

m = 800 kg, h = 12 m, g = 9.8 m/s² 

Formula: 

P.E = m g h

= 800 × 9.8 × 12

= 94080 J

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Problem: A 50 W electric bulb is used for 8 hours every day. How much electricity does it consume each day?

Solution:

Given: 

P = 50 W = 0.05 kW, t = 8 h

Energy consumed = Power × Time

Energy = 0.05 × 8

= 0.4 kWh 

or 0.4 units

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