The work done on an object equals its change in kinetic energy:

W = ΔKE

This combines ideas from kinematics, force and energy.

Example Question:

A giant 6.0 kg air hockey puck is sliding on a frictionless surface at a velocity of 2.0 m/s. Then a person pushes on it until it speeds up to 5.0 m/s.
a) How much work did the person do?
b) If the person pushed the puck over a distance of 4.0 m, how much force did they apply over that distance?

Example Solution:
a) Calculate Work
Initial KE = ½mv₀² = ½(6.0 kg)(2.0 m/s)² = 12 J
Final KE = ½mv² = ½(6.0 kg)(5.0 m/s)² = 75 J
W = ΔKE = (75 J − 12 J) = 63 J

b) Calculate Force
W = Fd
63 J = F (4.0 m)
F = 15.75 N

Proof

Let’s start with ΔKE and prove that it equals W.

ΔKE
½mv² − ½mv₀²
½m(v² – v₀²)
½m(v − v₀)(v + v₀)

Δv = v − v₀
v = (v − v₀)/2

Combining the previous three expressions, we get this:

mΔvv

Combine that with this:

Δv = aΔt

That yields this:

m(aΔt)v

Rearrange the order a bit:

(ma)(vΔt)

Now we can make these substitutions:
F = ma   (first parentheses above)
Δx = vΔt  (second parentheses above)

That yields this:

FΔx

This equals work.
W = FΔx
(Usually this is written as W = Fd, which means the same thing)

So, we started from ΔKE and manipulated equations until we proved that Work = ΔKE.