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Answer:

(a) The distance up the slope the wagon moves before coming to rest is approximately 21.74 m

(b) The distance the wagon comes to rest from the starting point is approximately 12.06 m

(c) The value of 'U' at which the wagon should be propelled if it is to come finally to rest at its starting point is approximately 3.214 m/s (the difference in value can come from calculating processes)

Explanation:

The wagon motion parameters are;

The mass of the wagon, m = 7,200 kg

The initial velocity with which the wagon is projected along the horizontal rail, v = U

The length of the horizontal portion of the rail = 100 m

The angle of inclination of the inclined portion of the rail, θ = sin⁻¹(0.01)

The exerted frictional resistance to motion of the rail, [tex]F_f[/tex] = 140 N

∴ θ = sin⁻¹(0.01)

The work done by the frictional force on the horizontal portion of the rail = 140 N × 100 m = 14,000 J

(a) If U = 3 m/s, we have;

Kinetic energy = 1/2·m·v²

The initial kinetic energy of the wagon, K.E. is given with the known parameters as follows;

K.E. = 1/2 × 7,200 kg × (3 m/s)² = 32,400 J

The energy, E, required to move a distance, 'd', up the slope is given as follows;

E = [tex]F_f[/tex] × d + m·g·h

Where;

[tex]F_f[/tex] = The friction force = 140 N

m = The mass of the wagon = 7,200 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = The height reached = d × sin(θ) = d × 0.01

Therefore;

E = 140 N × d₁ + 7,200 kg × 9.81 m/s² × d₁ × 0.01 = 846.32 N × d

The energy, [tex]E_{NET \ horizontal}[/tex], remaining from the horizontal portion of the rail is given as follows;

[tex]E_{NET \ horizontal}[/tex] = Initial kinetic energy of the wagon - Work done on frictional resistance on the horizontal portion of the rail

∴ [tex]E_{NET \ horizontal}[/tex] = 32,400 J - 14,000 J = 18,400 J

[tex]E_{NET \ horizontal}[/tex] = 18,400 J

Therefore, for the wagon with energy, [tex]E_{NET \ horizontal}[/tex] to move up the train, we get;

[tex]E_{NET \ horizontal}[/tex] = E

∴ 18,400 J = 846.32N × d

d₁ = 18,400 J/(846.36 N) ≈ 21.7401579 m

d₁ ≈ 21.74 m

The distance up the slope the wagon moves before coming to rest, d₁ ≈ 21.74 m

(b) Given that the initial velocity of the wagon, U = 3 m/s, the distance up the slope the wagon moves before coming to rest is given above as d₁ ≈ 21.74 m

The initial potential energy, PE, of the wagon while at the maximum height up the slope is given as follows;

P.E. = m·g·h = 7,200 kg × 9.81 m/s² × 21.74 × 0.01 m = 15,355.3968 J

The work done, 'W', on the frictional force on the return of the wagon is given as follows;

W = [tex]F_f[/tex] × d₂

Where d₂ = the distance moved by the wagon

By conservation of energy, we have;

P.E. = W

∴  15,355.3968 = 140 × d₂

d₂ = 15,355.4/140 = 109.681405714

Therefore;

The distance the wagon moves from the maximum height, d₂ ≈ 109.68 m

The distance the wagon comes to rest from the starting point, d₃, is given as follows;

d₃ = Horizontal distance + d₁ - d₂

d₃ = 100 m + 21.74 m - 109.68 m ≈ 12.06 m

The distance the wagon comes to rest from the starting point, d₃ ≈ 12.06 m

(c) For the wagon to come finally to rest at it starting point, we have;

The initial kinetic energy = The total work done

1/2·m·v² = 2 × [tex]F_f[/tex] × d

∴ 1/2 × 7,200 × U² = 2 × 140 × d₄

d₄ = 100 + (1/2·m·U² - 140×100)

(1/2·m·U² - 140×100)/(m·g) = h = d₁ × 0.01

∴ d₁ = (1/2·m·U² - 140×100)/(m·g×0.01)

d₄ = 100 + d₁

∴ d₄ = 100 + (1/2·m·U² - 140×100)/(m·g×0.01)

∴ 1/2 × 7,200 × U² = 2 × 140 × (100 + (1/2 × 7,200 × U² - 140×100)/(7,200 × 9.81 ×0.01))

3,600·U² = 280·(100 + (3,600·U² - 14,000)/706.32)

= 28000 + 280×3,600·U²/706.32 - 280 × 14,000/706.32

= 28000 - 280 × 14,000/706.32 + 1427.11518858·U²

3,600·U² - 1427.11518858·U² = 28000 - 280 × 14,000/706.32

U²·(3,600 - 1427.11518858) = (28000 - 280 × 14,000/706.32)

U² = (28000 - 280 × 14,000/706.32)/(3,600 - 1427.11518858) = 10.3319363649

U = √(10.3319363649) = 3.21433295801

The value of 'U' at which the wagon should be propelled if it is to come finally to rest at its starting point is U ≈ 3.214 m/s

Percentage error = (3.214-3.115)/3.214 × 100 ≈ 3.1% < 5% (Acceptable)

The difference in value can come from difference in calculating methods

Yeah order for a firm voltage dividers to operate properly The load resistance value should be at least Times greater than resistance value of the voltage divider bleeder resistor

Question 1: What is the power observed in the energy analyzer when the rated voltage(U1) is applied to the primary of the transformer, and there is no load at the secondary?
Question 2: Find the transformation ratio of the transformer using the values U1,U2 recorded in the experiment.
Question 3: Sketch the no-load operation graph of the transformer using the values U1, I2 and the values read in the energy analyzer.
Question 4: How can we find the number of turns of transformer?
Question 5: Explain the operation principle of the transformer.
Question 6: State your final observations about the experiment.

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