Why does work equal change in kinetic energy?

[jaydnul]

If I hit a solid brick wall with a 2kg hammer moving 2m/s, the initial KE would be 4J. Which means the work done would be 4J. But W=F•d says the work is 0 since the hammer comes to a complete stop at the wall without any displacement. ?

 

[NascentOxygen]

The wall deforms, absorbing some of the energy. Sure, d is small, but not zero; and F is large, so the product can be large. The hammer head deforms, too, absorbing energy like a spring being compressed.

In the situation you paint, I think the hammer bounces back with equal energy, so watch out for the recoil!

Whatever way you look at it, energy is conserved.

 

[CWatters]

But W=F•d says the work is 0 since the hammer comes to a complete stop at the wall without any displacement.

What NascentOxygen said.

Your statement isn't true. The hammer does not stop instantly without displacement. If it did the acceleration (actually deceleration) would be infinite which isn't possible.

The wall and/or the hammer will deform slightly. There is plenty of evidence for that... for example a sound wave will propagate through the wall. For that to happen there must be at least some distortion/compression of the wall.

Some of the energy may go into heating the wall or the sound waves but overall it's still conserved.