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:

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.

kasliwalalvi24 kasliwalalvi24 · Answer 2 · 2021-12-29T08:33:55+01:00

The mechanical energy is constant if the subject is an isolated system pertaining to conservative forces. The change in the thermal energy is 43,750 J.

Given that:

Mass = 5 kg

Initial velocity = 200 m/s

Final velocity = 150 m/s

Change in thermal energy = ?

The principle of conservation of mechanical energy states that it is equivalent to the sum of kinetic energies and potential energies in an isolated system, such that:

PE₁ + KE₁ = PE₂ + KE₂

As we know, the projectile fired from the ground level. The potential energy is zero.

The equation becomes:

[tex]\Delta KE &= KE_1 - KE_2\\\\\Delta KE &= \dfrac{1}{2} (mv_1)^2-\dfrac{1}{2} (mv_2)^2\\\\\Delta KE &= \dfrac{1}{2}[v_1^{(2)} - v_2^{(2)}]\\\\\Delta KE &= \dfrac{1}{2} \times 5 (200^2 - 150^2)[/tex]

[tex]\Delta KE &= 43,750 J[/tex]

Thus, the change in the thermal energy is 43,750 J.

To know more about thermal energy, refer to the following link:

https://brainly.com/question/11278589