Explanation:
To propose the solution of this problem we must use the relationship between work and energy
W = ΔK
the change in kinetic energy is
ΔK = K_f -K₀
ΔK = ½ m v_f² - ½ m v₀² = ½ m (v_f² - v₀²)
in this case
ΔK =
we can find work with the first law of thermodynamics
= Q + W
where \Delta E_{int} is the internal energy of the body, usually measured in the form of an increase in the temperature of the system
W = \Delta E_{int} - Q
if we consider that the internal energy does not change
W = -Q
we substitute everything in the first equation
-Q =
Because they are squared, the variables are positive, therefore, for the equation to be fulfilled, the exit velocity must be less than the entrance velocity.