A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25°above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip the change in the thermal energy of the projectile and air is:

Given Information:

Mass = m = 5 kg

Initial velocity = v₁ = 200 m/s

Final velocity = v₂ = 150 m/s

Required Information:

change in thermal energy = ?

Answer:

ΔKE = 43,750 J

Explanation:

According to the principle of conservation of mechanical energy

PE₁ + KE₁ = PE₂ + KE₂

The projectile is fired from ground level so there is zero energy stored due tot he position of the projectile which means that the change in potential energy is zero, so the equation becomes

ΔKE = KE₁ - KE₂

ΔKE = ½mv₁² - ½mv₂²

ΔKE = ½m (v₁² - v₂²)

ΔKE = ½*5 (200² - 150²)

ΔKE = 43,750 J

Therefore, the change in the thermal energy of the projectile and air is 43,750 Joules.

the change in thermal energy of the projectile is 43.8 kJ

Explanation:

Given;

mass of the object, m = 5kg

initial velocity of the projectile, v₁ = 200 m/s

final velocity of the projectile, v₂ = 150 m/s

To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.

Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²

KE = Δ¹/₂mv² = ¹/₂m (v₁²-v₂²)

KE = ¹/₂ * 5 (200²-150²) = 2.5 (17500) = 43750 J = 43.8 kJ

Therefore, the change in thermal energy of the projectile is 43.8 kJ