The Math Behind the Work-Energy Theorem

Previously we saw how Santa and his reindeer team encountered a wind gust which generated enough force to slow them from an initial velocity of v₁ to a final velocity, v_2, over a distance, d.   here we’ll begin using the Work-Energy Theorem to see if Santa was able to keep to his Christmas delivery schedule and get all the good boys and girls their gifts in time.

Before we can work with the Work-Energy Theorem, we must first revisit the formula it’s predicated upon, de Coriolis’ formula for kinetic energy,

KE = ½ × m × v²                            (1)

where, KE is kinetic energy, m is the moving object’s mass, and v its velocity.

The equation behind the Work-Energy Theorem is,

W = KE₂ – KE₁                                        (2)

where W is the work performed, KE₁ is the moving object’s initial kinetic energy and KE₂its final kinetic energy after it has slowed or stopped.   In cases where the object has come to a complete stop KE₂ is equal to zero, since the velocity of a stationary object is zero.

In order to work with equation (2) we must first expand it into a more useful format that quantifies an object’s mass and initial and final velocities.   We’ll do that by substituting equation (1) into equation (2).   The result of that term substitution is,

W = [½ × m × v₂² ] – [½ × m × v₁²]      (3)

Factoring out like terms, equation (3) is simplified to,

W = ½ × m × [v₂² – v₁²]                      (4)

Now according to de Coriolis, work is equal to force, F, times distance, d.   So substituting these terms for W in equation (4), the expanded version of the Work-Energy Theorem becomes,

F × d = ½ × m × [v₂² – v₁²]                 (5)