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A 5 kg projectile is fired at an angle of 25o above the horizontal. Its initial velocity is 200 m/s and just before it hits the ground its velocity is 150 m/s. What is the change in the mechanical energy of the projectile?

User Declension
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1 Answer

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Answer:

The change in the mechanical energy of the projectile is 43,750 J

Step-by-step explanation:

Given;

mass of the projectile, m = 5 kg

initial velocity of the projectile, u = 200 m/s

final velocity of the projectile, v = 150 m/s

The change in mechanical energy is calculated from the principle of conservation of energy;

ΔP.E = ΔK.E

The change in potential energy is zero (0)

0 = ΔK.E

ΔK.E = K.E₁ - K.E₂

ΔK.E = ¹/₂mu² - ¹/₂mv²

ΔK.E = ¹/₂m(u² - v²)

ΔK.E = ¹/₂ x 5(200² - 150²)

ΔK.E = 43,750 J

Therefore, the change in the mechanical energy of the projectile is 43,750 J

User Natemcmaster
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