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SSSSS [86.1K]
4 years ago
15

A 1500 kg car on flat ground is moving 5.25 m/s. Its engine creates a 1250 N forward force as the car moves 42.8 m. What is its

final velocity?​

Physics
1 answer:
kari74 [83]4 years ago
5 0

Answer:

9.93 m/s

Explanation:

The following data were obtained from the question:

Mass (m) = 1500 Kg

Initial velocity (u) = 5.25 m/s

Force (F) = 1250 N

Distance (s) = 42.8 m

Final velocity (v) =...?

Next, we shall determine the acceleration of car. This can be achieved as shown below:

Force (F) = mass (m) x acceleration (a)

1250 = 1500 x a

Divide both side by 1500

a = 1250/1500

a = 0.83 m/s²

Therefore, the acceleration of the car is 0.83 m/s²

Now, we can determine the final velocity of car as shown below:

Initial velocity (u) = 5.25 m/s

Distance (s) = 42.8 m

Acceleration (a) = 0.83 m/s²

Final velocity (v) =...?

v² = u² + 2as

v² = (5.25)² + 2 x 0.83 x 42.8

v² = 27.5625 + 71.048

v² = 98.6105

Take the square root of both side

v = √98.6105

v = 9.93 m/s

Therefore, the final velocity of the car is 9.93 m/s

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Explanation:

From the law of conservation of energy,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

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Given that (E₂ - E₁) = change in internal energy = ΔE = 1/2ΔK where ΔK = change in kinetic energy. So, ΔE = 1/2ΔK = 1/2(K₂ - K₁) = K₂/2 = 1/2(m + M)v²/2 = (m + M)v²/4

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taking square root of both sides, we have

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Substituting the values of the variables into the equation, we have

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v = √[13.8422 kgm/s²/{9.432 kg)}]

v = √(1.4676 m²/s²)

v = 1.21 m/s

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