WORK, POWER AND ENERGY – PART 2

This article is formulated according to the 6th chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.

∴ 𝑷𝒂𝒗 = 𝑾/𝒕 = 𝑬/𝒕

Instantaneous power

It is the limiting value of the average power as the time interval approaches zero. We have, 𝑃 = lim 𝑑→0 π‘ƒπ‘Žπ‘£

𝑷 = 𝒅𝑾/𝒅𝒕

But π‘‘π‘Š = 𝐹⃗ βˆ™ 𝑑𝑠⃗ where 𝑑𝑠⃗ β†’ diplacment

∴ 𝑃 = 𝐹⃗ . (𝑑𝑠⃗/𝑑𝑑)

𝑷 = 𝑭⃗ βˆ™ 𝒗⃗

Note: If the Force 𝐹 acts in the direction of motion then πœƒ = 0, cos 0 = 1 ∴ 𝑃 = 𝐹𝑣 cos πœƒ = 𝐹𝑣 cos 0

𝑷 = 𝑭𝒗

Unit of Power

SI unit of power is watt (W), its dimensions are [ML2T-3] and 1 π‘€π‘Žπ‘‘π‘‘ = 1J/1𝑆

Watt

The power is said to be 1 watt of one-joule work is done/energy is consumed in one second by any agent.

Practical unit

The practical unit of power is horsepower (hp). 1 horsepower (hp) = 746 W

Collisions

The term collision refers to the interaction between two bodies due to which the direction and magnitude of the velocity of the colliding bodies change. In a collision, the external force acting on the colliding particles is neglected.

Types of collisions

Elastic collisions

In a collision, if both the linear momentum and kinetic energy of the system are conserved then such a collision is called an elastic collision. It means the linear momentum and the kinetic energy of the system before and after the collision is the same.

Ex:

(i) Perfectly elastic collision is a rare event, collisions between atomic particles are nearly elastic.

(ii) A ball dropped from a certain height will rebound to the same height if the collision with the surface is elastic.

Inelastic (plastic) collision

A collision is said to be inelastic if the linear momentum of the system remains conserved, but its kinetic energy is not conserved. In an inelastic collision, the loss of kinetic energy appears in the form of heat, elastic potential energy sound, and light energy.

Ex: Collisions between macroscopic bodies are inelastic A ball dropped from a certain height will not rebound to some height if the collision with the surface is inelastic.

Perfectly inelastic collision

A collision is said to be perfectly inelastic if the two bodies after collision stick together and move as one body.

Ex: When a moving bullet hits the stationary wooden block, then it is embedded into the wooden block, and both move as one body.

Difference between elastic and inelastic collisions

Elastic collisionInelastic collision
Both the momentum and kinetic energy are conservedMomentum is conserved but the kinetic energy is not conserved
Force involved in the collision are conservativeForces involved are non-conservative
Mechanical energy is not converted into other forms of energyMechanical energy is converted into other forms of energy

Collisions in one Dimension

If the initial velocities and final velocities of both bodies are along the same straight line, then it is called a one-dimensional collision or Head-on collision.

Expression for loss in Kinetic energy in a completely inelastic collision

Consider two masses m1 and m2. The particle m1 is moving with speed u1 and m2 is at rest. After collision both masses m1 and m2 stick to each other and move as one body with velocity 𝑣. In this type of collision, the linear momentum is conserved.

π‘š1𝑒1 + π‘š2𝑒2 = (π‘š1 + π‘š2)𝑣

π‘š1𝑒1 = (π‘š1 + π‘š2)𝑣  (𝑣2 = 0 β‡’ π‘š2𝑒2 = 0)

𝑣 = (π‘š1𝑒1)/(π‘š1 + π‘š2)

As the final kinetic energy is less than the initial kinetic energy, The change in kinetic energy is,

βˆ†πΎ = (1/2)π‘šπ‘’12 – (1/2)(π‘š1 + π‘š2)𝑣2

Substituting for 𝑣 from the above equation

βˆ†πΎ = (1/2)π‘šπ‘’12 – [(1/2)(π‘š1 + π‘š2)(π‘š12/(π‘š1 + π‘š2)2𝑒12)]

βˆ†πΎ = (1/2)π‘š1𝑒12 – (1/2)(π‘š12/(π‘š1 + π‘š2))𝑒12

βˆ†πΎ = (1/2)π‘š1𝑒12[1 – (π‘š1/(π‘š1 + π‘š2)]

βˆ†πΎ = (1/2)π‘š1𝑒12[(π‘š1 + π‘š2 βˆ’ π‘š1)/(π‘š1 + π‘š2)]

βˆ†πΎ = (1/2)π‘š1𝑒12[π‘š2/(π‘š1 + π‘š2)]

βˆ†π‘² = (𝟏/𝟐)[(π’ŽπŸπ’ŽπŸ/(π’ŽπŸ + π’ŽπŸ)]π’–πŸπŸ

This loss in kinetic energy appears as the sound and heat energies, Thus total energy is conserved.

Expression for final velocities of the bodies in an elastic collision

Consider two masses m1 and m2. The body m1 is moving with speed u1 and m2 is at rest. After the collision, their velocities are v1 and v2 respectively, and are moving in the same straight line. As the collision is elastic both the linear momentum and kinetic energies are conserved.

π‘€π‘œπ‘šπ‘’π‘›π‘‘π‘’π‘š β‡’ π‘š1𝑒1 + π‘š2𝑒2 = π‘š1𝑣1 + π‘š2𝑣2

π‘š1𝑒1 = π‘š1𝑣1 + π‘š2𝑣2 (π‘š2𝑒2 = 0)

π‘š1𝑒1 βˆ’ π‘š1𝑣1 = π‘š2𝑣2

π‘š1(𝑒1 βˆ’ 𝑣1) = π‘š2𝑣2 —-Γ  (1)

Kinetic energy β‡’ (1/2)π‘š1𝑒12 = (1/2)π‘š1𝑣12 + (1/2)π‘š1𝑣22

π‘š1𝑒12 = π‘š1𝑣12 + π‘š2𝑣22

π‘š1𝑒12 βˆ’ π‘š1𝑣12 = π‘š2𝑣22

π‘š1(𝑒12 βˆ’ 𝑣12) = π‘š2𝑣22 —-Γ  (2)

Dividing equation (2) by equation (1)

π‘š1(𝑒12 βˆ’ 𝑣12)/π‘š1(𝑒1 βˆ’ 𝑣1) = π‘š2𝑣22/π‘š2𝑣2

π‘š1(𝑒1 + 𝑣1)(𝑒1 βˆ’ 𝑣1)/π‘š1(𝑒1 βˆ’ 𝑣1) = π‘š2𝑣22/π‘š2𝑣2

𝑒1 + 𝑣1 = 𝑣2

𝑣2 = 𝑒1 + 𝑣1 βˆ’ βˆ’ βˆ’ βˆ’ βˆ’> (3)

Substituting equation (3) in equation π‘š1𝑒1 βˆ’ π‘š1𝑣1 = π‘š2𝑣2

π‘š1𝑒1 βˆ’ π‘š1𝑣1 = π‘š2(𝑒1 + 𝑣1)

π‘š1𝑒1 βˆ’ π‘š1𝑣1 = π‘š2𝑒1 + π‘š2𝑣1

π‘š1𝑒1 βˆ’ π‘š2𝑒1 = π‘š1𝑣1 + π‘š2𝑣1

𝑒1(π‘š1 βˆ’ π‘š2) = 𝑣1(π‘š1 + π‘š2)

π’—πŸ = [(π’ŽπŸ βˆ’ π’ŽπŸ)/(π’ŽπŸ + π’ŽπŸ)]π’–πŸ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’> (4)

Substituting (4) in (3)

𝑣2 = 𝑒1 + ((π‘š1 βˆ’ π‘š2)/(π‘š1 + π‘š2))𝑒1

𝑣2 = 𝑒1[1 + ((π‘š1 βˆ’ π‘š2)/(π‘š1 + π‘š2))]

𝑣2 = 𝑒1[(π‘š1 + π‘š2 + π‘š1 βˆ’ π‘š2)/(π‘š1 + π‘š2)]

𝑣2 = 𝑒1[2π‘š1/(π‘š1 + π‘š2)]

π’—πŸ = (πŸπ’ŽπŸ/(π’ŽπŸ + π’ŽπŸ))π’–πŸ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’> (5)

Equation (4) and (5) gives the expression for the final velocities of the masses m1 and m2 respectively. 𝑖. 𝑒.,

𝑣1 = [(π‘š1 βˆ’ π‘š2)/(π‘š1 + π‘š2)]𝑒1 and 𝑣2 = [2π‘š1/(π‘š1 + π‘š2)]𝑒1

Special cases: Case (1): If two masses one equal. 𝑖. 𝑒.,

π‘š1 = π‘š2 = π‘š

𝑣1 = [(π‘š – π‘š)/(π‘š + π‘š)]𝑒1

𝑣1 = 0.

The final velocity of mass m1 will become zero. It means after collision mass m1 comes to rest and pushes off the second mass m2.

𝑣2 = [2π‘š/(π‘š + π‘š)]𝑒1 = (2π‘š/2π‘š)𝑒1

𝑣2 = 𝑒1

The final velocity of mass m2 = Initial velocity of mass m1.

It means the second mass m2 moves with a velocity of m1

Case (2): If one the mass dominates. If m2 >> m1

𝑣2 = [(π‘š1 βˆ’ π‘š2)/(π‘š1 + π‘š2)]𝑒1 π‘Žπ‘  π‘š2 > π‘š1 then π‘š1 βˆ’ π‘š2 = βˆ’π‘£π‘’

𝑣1 = βˆ’π‘’1

The final velocity of mass m1 gets reversed which means it comes back with the same velocity.

𝑣2 = [2π‘š1/(π‘š1 + π‘š2)]𝑒1

𝑣2 = 2π‘š1𝑒1/π‘š2 β‰ˆ 0

(𝑏𝑒𝑐a𝑒𝑠𝑒 π‘š2 ≫ π‘š1, 2π‘š1𝑒1/π‘š2 β‰ˆ 0)

𝑣2 = 0

Heavy mass m2 practically remains at rest.

Case (3): If π‘š1 ≫ π‘š2

𝑣1 = [(π‘š1 βˆ’ π‘š2)/(π‘š1 + π‘š2)]𝑒1

𝑣1 = (π‘š1/π‘š1)𝑒1

(π‘š2 π‘π‘Žπ‘› 𝑏𝑒 𝑛𝑒𝑔𝑙𝑒𝑐𝑑𝑒𝑑)

𝑣1 = 𝑒1

The mass m1 moves with the same velocity.

𝑣2 = [2π‘š1/(π‘š1 + π‘š2)]𝑒1 = (2π‘š1/π‘š1)𝑒1

𝑣2 = 2𝑒1

The lighter mass m2 moves with twice the velocity of the heavy body.

Collisions in two Dimensions

Consider two masses m1 and m2. The body m1 is moving with speed u1 and m2 is at rest. The mass m1 collides with the stationary mass m2 and this is shown in the figure. After the collision, the masses m1 and m2 fly off in different directions. The body m1 moves with velocity v1 making an angle πœƒ1, called deflecting angle, with π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠 and mass m2 move with a velocity v2 making an angle πœƒ2, called the angle of recoil, with π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠. In all types of collision, momentum is conserved. Therefore, the conservation of momentum along π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠 and 𝑦 βˆ’ π‘Žπ‘₯𝑖𝑠 gives,

(i) π‘š1𝑒1 = π‘š1𝑣1 cos πœƒ1 + π‘š2𝑣2 cos πœƒ2 along π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠

(ii) 0 = π‘š1𝑣2 sin πœƒ1 βˆ’ π‘š2𝑣2 sin πœƒ2 along 𝑦 βˆ’ π‘Žπ‘₯𝑖𝑠

π‘š1𝑣2 sin πœƒ1 = π‘š2𝑣2 sin πœƒ2

If the collision is perfectly elastic, then kinetic energy is also conserved.

(iii) (1/2)π‘š1𝑒12 = (1/2)π‘š1𝑣12 + (1/2)π‘š2𝑣22

Assume that π‘š1, π‘š2, and 𝑒1 are known. Now the motion after collision involves four unknowns i.e., 𝑣1, 𝑣2, πœƒ1, and πœƒ2. To evaluate these four quantities, we need a fourth equation. However, we have only three equations. Therefore, the fourth equation is to be developed, but this process of developing the fourth equation is quite complicated. To overcome this problem the easiest way of developing the fourth equation is the measure the angle of defection πœƒ1 and the angle of recoil πœƒ2 experimentally.

Problems

1. Show that 𝐴 = 2𝑖̂+ 𝑗̂+ 2π‘˜Μ‚ and 𝐡 = 2π‘–Μ‚βˆ’ 2π‘—Μ‚βˆ’ π‘˜Μ‚ are perpendicular to each other.

𝐴 βˆ™ 𝐡 = 𝐴π‘₯𝐡π‘₯ + 𝐴𝑦𝐡𝑦 + 𝐴𝑧𝐡𝑧

𝐴 βˆ™ 𝐡 = (2)(2) + (1)(βˆ’2) + (2)(βˆ’1) = 4 βˆ’ 2 βˆ’ 2 = 0

𝐴 . 𝐡 = 0, So, vectors are perpendicular to each other.

2. Find the angle between two vectors 𝐴 = 𝑖̂+ 2𝑗̂+ π‘˜Μ‚ and 𝐡 = 2π‘–Μ‚βˆ’ 𝑗̂+ π‘˜Μ‚.

𝐴 βˆ™ 𝐡 = (1)(2) + (2)(βˆ’1) + (1)(1) = 2 βˆ’ 2 + 1 = 1

|𝐴| = √[(1)2 + (2)2 + (1)2] = √(1 + 4 + 1) = √6

|𝐡| = √[(2)2 + (βˆ’1) 2 + (1)2] = √(4 + 1 + 1) = √6

πœƒ = π‘π‘œπ‘ βˆ’1 [𝐴 . 𝐡/(|𝐴||𝐡|)] = π‘π‘œπ‘ βˆ’1[1/(√6 Γ— √6)]

πœƒ = π‘π‘œπ‘ βˆ’1(1/6) = π‘π‘œπ‘ βˆ’1 0.1667

πœƒ = 80Β°24’

3. Find the angle between force 𝐹 = (3𝑖̂+ 4π‘—Μ‚βˆ’ 5π‘˜Μ‚) unit and displacement 𝑑 = (5𝑖̂+ 4𝑗̂+ 3π‘˜Μ‚ unit.

𝐹 βˆ™ 𝑑 = (3)(5) + (4)(4) + (βˆ’5)(3) = 15 + 16 βˆ’ 15 = 16 units

|𝐹| = √(32 + 42 + (βˆ’5)2) = √(9 + 16 + 25) = √50 unit

|𝑑| = √(52 + 42 + 32) = √(25 + 16 + 9) = √50 units

cos πœƒ = 𝐹 . 𝑑 /|𝐹||𝑑|

cos πœƒ = 16 /(√50 Γ— √50) = 16/50

cos πœƒ = 0.32

πœƒ = π‘π‘œπ‘ βˆ’1(0.32) β‰ˆ 71Β°

4. A body constrained to move along the z-axis of a coordinating system is subjected to a constant force 𝐹 given by 𝐹 = βˆ’π‘–Μ‚+ 2𝑗̂+ 3π‘˜Μ‚ 𝑁, where 𝑖,̂𝑗̂ and π‘˜Μ‚ are unit vectors along x,y and z – axis of the system respectively. What is the work done by this force in moving the body a distance of 4m along the z-axis.

𝐹 = βˆ’π‘–Μ‚+ 2𝑗̂+ 3π‘˜Μ‚ 𝑁

π‘Š = 𝐹 βˆ™ 𝑑 = 𝐹π‘₯𝑑π‘₯ + 𝐹𝑦𝑑𝑦 + 𝐹𝑧𝑑𝑧

𝑑 = 4π‘˜π‘š

π‘Š = (βˆ’π‘–Μ‚+ 2𝑗̂+ 3π‘˜Μ‚) βˆ™ (0𝑖̂+ 0𝑗̂+ 4π‘˜Μ‚)

π‘Š = (βˆ’1)(0) + (2)(0) + (3)(4) = 12 𝐽

5. In a ballistics demonstration a police officer fires a bullet of mass 50g with speed 200ms-1 on soft plywood of thickness 2 cm. The bullet emerges with only 10% of its initial kinetic energy what is the emergent speed of the bullet?

π‘š = 50 Γ— 10βˆ’3π‘˜π‘”

Initial kinetic energy is π‘˜π‘– = (1/2)π‘šπ‘£π‘–2

𝑣𝑖 = 200π‘šπ‘ βˆ’1

π‘˜π‘– = (1/2) Γ— 50 Γ— 100 Γ— 200 Γ— 10βˆ’3

π‘˜π‘– = 1000000 Γ— 10βˆ’3 = 1000𝐽

Final kinetic energy is 10% of initial kinetic energy.

π‘˜π‘“ = 10% π‘˜π‘– = (10/100) Γ— 1000 = 100𝐽

𝑣𝑓2 = 2π‘˜π‘“/π‘š

𝑣𝑓2 = (2 Γ— 100)/(50 Γ— 103) = 4 Γ— 103

𝑣𝑓 = √(4 Γ— 103) = 63.2π‘šπ‘ βˆ’1

6.  An electron and proton are detected in a cosmic ray experiment, the first with kinetic energy 10π‘˜π‘’π‘‰, and the second with 100π‘˜π‘’π‘‰, which is faster, the electron or the proton? Obtain the ratio of their speeds?

π‘šπ‘’ = 9.11 Γ— 10βˆ’31π‘˜π‘”

π‘šπ‘ = 1.67 Γ— 10βˆ’27π‘˜π‘”

π‘˜π‘’ = 10 Γ— 103 Γ— 1.6 Γ— 10βˆ’19𝐽

𝐾𝑝 = 100 Γ— 103 Γ— 1.6 Γ— 10βˆ’19𝐽

Velocity of electron is,

𝑣𝑒 = √ (2 Γ— 𝐾𝑒/π‘šπ‘’)

𝑣𝑒 = [(2 Γ— 10 Γ— 103 Γ— 1.6 Γ— 10βˆ’19)/(9.11 Γ— 10βˆ’31)]1/2

𝑣𝑒 = 5.9267 Γ— 107π‘šπ‘ βˆ’1

Velocity of the proton,

𝑣𝑃 = √(2𝐾𝑃/π‘šπ‘ƒ)

𝑣𝑃 = [(2 Γ— 100 Γ— 103 Γ— 1.6 Γ— 10βˆ’19)/(1.67 Γ— 10βˆ’27]1/2

𝑣𝑃 = 4.3774 Γ— 106π‘šπ‘ βˆ’1

Comparing ve and vp we get electron is moving faster,

𝑣𝑒/𝑣𝑝 = (5.9267 Γ— 107)/(4.3774 Γ— 106) = 13.54

7. A rain drop of radius 2mm falls from a height of 500m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height it attains its maximum (terminal) speed and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire Journey if its speed on reaching the ground is 10ms-1.

π‘Ÿ = 2π‘šπ‘š = 2 Γ— 10-3π‘š

β„Ž1 = 250 π‘š

β„Ž2 = 250 π‘š

β„Ž = β„Ž1 + β„Ž2 = 500π‘š

Mass of the drop is mass = volume Γ— density

Volume = (4/3)πœ‹π‘Ÿ3 = (4/3) Γ— 3.14 Γ— (2 Γ— 10-3)3 = 3.3493 Γ— 10βˆ’8π‘˜π‘”/π‘š3

Mass = 3.3493 Γ— 10βˆ’8 Γ— 103 = 3.3493 Γ— 10βˆ’5π‘˜π‘”

π‘Šπ‘”1 =? π‘Šπ‘”1 = π‘šπ‘”β„Ž = 3.3493 Γ— 10βˆ’5 Γ— 9.8 Γ— 250 = 0.082 𝐽

π‘Šπ‘”2 =? π‘Šπ‘Ÿ =?

Since the distance travelled in the second half is same as first halt therefore π‘Šg2 = 0.082𝐽

Total work done on it is, π‘Š = π‘Šπ‘”1 + π‘Šπ‘”2 + π‘Šπ‘Ÿ

From work energy theorem, π‘Š = βˆ†πΎ

π‘Šπ‘”1 + π‘Šπ‘”2 + π‘Šπ‘Ÿ = (1/2)π‘šπ‘£π‘“2 – (1/2)π‘šπ‘£π‘–2

0.082 + 0.08 + Wr = [(1/2) Γ— 3.3493 Γ— 105 Γ— 102) βˆ’ (0) [∡ 𝑣𝑖 = 0]

0.164 + Wπ‘Ÿ = 1.6746 Γ— 10βˆ’3 π‘Šπ‘Ÿ = 1.6746 Γ— 10βˆ’3 βˆ’ 0.164 π‘Šπ‘Ÿ = βˆ’0.1623 𝐽

8. A body of mass 0.5 Kg travels in a straight line with velocity 𝑣 = π‘Žπ‘₯3/2, where π‘Ž = 5π‘šβˆ’1/2π‘ βˆ’1. What is the work done by the net force during its displacement from π‘₯ = 0 to π‘₯ = 2π‘š.

Initial velocity at π‘₯ = 0. π‘š = 0.5π‘˜π‘”. 𝑣𝑖 = π‘Žπ‘₯3/2 = 5 Γ— (0)3/2 = 0

Final velocity at π‘₯ = 2, 𝑣𝑓 = π‘Žπ‘₯3/2 = 5 Γ— (2)3/2 = 5 Γ— 2.8284 = 14.1421π‘šπ‘ βˆ’1

From work energy theorem, π‘Š = βˆ†πΎ = (1/2)π‘šπ‘£π‘“2 βˆ’ (1/2)π‘šπ‘£i2 = ((1/2) Γ— 0.5 Γ— 14.14212) = (0) ∡ 𝑣𝑖 = 0

π‘Š = 49.999 𝐽 β‰ˆ 50𝐽

9. The potential energy of a certain spring when stretched through a distance π‘₯ is 10J. What is the amount of work done on the same spring to stretch it through an additional distance?

We have (1/2)π‘˜π‘₯2 = 10𝐽

When π‘₯ becomes 2π‘₯, then, New P.E = (1/2)𝐾(2π‘₯)2 = (1/2)𝐾(4π‘₯2) = (1/2)𝐾 π‘₯2 Γ— 4 = 10 Γ— 4 = 40𝐽

Work done = 40 𝐽 βˆ’ 10𝐽 = 30𝐽

10. A 100W bulb is on for 10 hours – calculate the energy consumed in kWh. To find energy in kWh, power must be in kW and time in hours.

𝑃 = 100𝑀 = 100 Γ—π‘˜π‘Š

1000 = 100 Γ— 10βˆ’3π‘˜π‘Š ∴ 𝑃 = 𝐸/𝑑

t = 10 h

𝐸 = 𝑝𝑑 = 100 Γ— 10βˆ’3 Γ— 10 = 1000 Γ— 10βˆ’3

𝐸 = 1000/103 = 1 π‘˜π‘€β„Ž

11. A pump on the ground floor of a building can pump up water to fill a tank of volume 30m3 in 15 min. If the tank is 40m above the ground and the efficiency of the pump is 30% how much electric power is consumed by the pump? (When water is pumped up against the gravitational force, work has to be done, and this work done is stored as potential energy.)

Volume 𝑣 = 30π‘š3

Density of water 𝑓 = 1000 π‘˜π‘”/π‘š3 ∴ Mass of the water, m = volume Γ— density

π‘š = 30 Γ— 1000 = 3 Γ— 104π‘˜π‘”

∴ Work done = π‘šπ‘”β„Ž = 3 Γ— 104 Γ— 9.8 Γ— 40

Work done = 1176 Γ— 104 𝐽

To do the above work actual power needed is π‘ƒπ‘Žπ‘ = π‘Š/𝑑 = (1176Γ—104)/(15Γ—60) = 13.0667 Γ— 103 watt

But the efficiency of the motor is only 30% therefore the power required is

πœ‚ = (π‘Žπ‘π‘‘π‘’π‘Žπ‘™ π‘π‘œπ‘€π‘’π‘Ÿ/𝑠𝑒𝑝𝑝𝑙𝑖𝑒𝑑 π‘π‘œπ‘€π‘’π‘Ÿ) Γ— 100

Supplied power = (π‘Žπ‘π‘‘π‘’π‘Žπ‘™ π‘π‘œπ‘€π‘’π‘Ÿ/πœ‚) Γ— 100

Supplied power = (13 Γ— 0.667 Γ— 103/30) Γ— 100

Supplied power = 43555.5556 π‘Š

12. A family uses 8kw of power. Direct solar energy is incident on the horizontal surface at an average rate of 200W/m2. It 20% of this energy can be converted to useful electrical energy how large an area is needed to supply 8 kW?

Power used by the family = 8π‘˜π‘Š = 8000 π‘Š

But it is only 20% of the total solar power supplied.

∴ 20% of solar power = 8000N

Solar power supplied = (8000/20) Γ— 100 = 40000π‘Š

Now, Area of the surface receives power of 200W = 1π‘š2

Area of the surface receives 1π‘Š power = 1π‘š2/200

Area of the surface receives 40000π‘Š power is = (1/200) Γ— 40000 = 200π‘š2

∴ Area needed = 200π‘š2

Area needed is nearly equal to 14π‘š Γ— 14π‘š roof.

13. The blades of a wind – mill sweep out a circle of area β€˜A’ (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time β€˜t’ ? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the winds energy into electrical energy and that A = 30m2,v= 36 km/h and the density of the air is 1.2kgm-3 what is the electrical power produced?

(a) Volume of air passing per see = Area Γ— velocity = Av

Mass of the wind passing per sec = volume Γ— density = 𝐴𝑣 Γ— 𝜌 = π΄π‘£πœŒ

Mass of the wind passing in time 𝑑 = π΄π‘£πœŒπ‘‘

(b) Kinetic energy of wind (Air) = (1/2)π‘šπ‘£2

Kinetic energy of wind (Air) = (1/2)(π΄π‘£πœŒπ‘‘)𝑣2

Kinetic energy of wind (Air) = (1/2)π΄πœŒπ‘£3𝑑

(c) The wind mill converts only 25% of the wind energy

∴ Energy available is = 25% of KE

Energy available = (25/100) Γ— (1/2)π΄πœŒπ‘£3𝑑 = (1/8)π΄πœŒπ‘£3𝑑

∴ π‘ƒπ‘œπ‘€π‘’π‘Ÿ π‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘’π‘‘ = 𝐸𝑛eπ‘Ÿπ‘”π‘¦/π‘‘π‘–π‘šπ‘’ = (1/8)π΄πœŒπ‘£3𝑑/𝑑 π‘ƒπ‘œπ‘€π‘’π‘Ÿ π‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘’π‘‘ = π΄πœŒπ‘£3/8 = (30 Γ— 1.2 Γ— (10)3)/8

π‘ƒπ‘œπ‘€π‘’π‘Ÿ π‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘’π‘‘ = 4500π‘Š = 4.5 π‘˜π‘Š

14. A neutron having mass of 1.67 Γ— 10βˆ’27kg and moving at 108π‘šπ‘ βˆ’1 collides with deuteron at rest and stick to it. Calculate the speed of the combination given mass of the deuteron is 3.34 Γ— 10βˆ’27kg.

Initial momentum = π‘šπ‘›π‘’π‘› + π‘šπ‘‘π‘’π‘‘ = 1.67 Γ— 10βˆ’27 Γ— 108 + 0 = 1.67 Γ— 10βˆ’19

Final momentum = (π‘šπ‘› + π‘šπ‘‘)𝑣 = (1.67 Γ— 10βˆ’27 + 3.34 Γ— 10βˆ’27)𝑣

As the momentum is conserved, Initial momentum = final momentum

1.67 Γ— 10βˆ’19 = 5.01 Γ— 10βˆ’27 Γ— 𝑣

𝑣 = (1.67 Γ— 1019)/(5.01 Γ— 10βˆ’27) = 3.3 Γ— 10βˆ’6π‘šπ‘ βˆ’1

Important Questions for Exams.

One mark.

1. State work-energy theorem.

2. When work is said to be done? or What is meant by β€œWork done by a force”? or Define work done by the force.

3. Write an expression for potential energy of a compressed spring.

4. Mention the expression for work done by a force in vector form.

5. What type of energy decreases when a body is falling freely?

6. What does the area under β€˜force-displacement’ curve represents?

7. What is potential energy?

8. What is meant by collision?

9. When the work done by a force does is zero?

10. Define power.

Two marks.

1. Distinguish between elastic and inelastic collision.

2. Give one example for elastic and inelastic collision.

3. What are elastic and inelastic collision?

Three marks.

1. What is the value of (a) 𝑖̂. 𝑖 Μ‚ (b) 𝑗̂.𝑗̂ (c) π‘˜Μ‚. π‘˜Μ‚

2. What is the value of (a) 𝑖̂.𝑗 Μ‚ (b) 𝑗̂. π‘˜Μ‚ (c) π‘˜Μ‚.𝑖̂

3. Give an example for (a) positive work (b) Negative work (c) zero work.

4. What is potential energy? Derive an expression for it.

5. Define kinetic energy. Give its unit and dimensional formula.

6. Obtain an expression for gravitational potential energy.

7. Define power. Show that the power is equal to the dot product of force and velocity.

8. Prove Work-Energy theorem for the motion of a particle under a constant acceleration. or Show that 𝐾𝑓 βˆ’ 𝐾𝑖 = π‘Š, Where 𝐾𝑓 and 𝐾𝑖 represent final and initial kinetic energies respectively and π‘Š represents work. or Prove the work-energy theorem in case of a rectilinear motion under constant acceleration. or Show that work done by the system is equal to change in energy.

9. Distinguish between Conservative force and Non-conservative force.

Five marks.

1. State the principle of conservation of energy and illustrate it in case of freely falling body.

2. State the law of conservation of mechanical energy. Show that the mechanical energy of a body dropped from a height is conserved. or State the law of conservation of mechanical energy. Show that the mechanical energy of a body falling freely under gravity is conserved. or State the law of conservation of mechanical energy and show that the mechanical energy is conserved in the case of a freely falling body. or State and prove the law of conservation of mechanical energy for a body falling under gravity. or Verify the law of conservation of mechanical energy in case of freely falling body.

3. State and Prove Work-Energy theorem for a constant force.

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