Push a box across the floor and let go: it slows and stops. Its kinetic energy is gone, but energy is conserved. It turned into thermal energy, warming both the box and the floor, by an amount you can calculate with Kinetic Friction.
Key idea
When kinetic friction $f_k$ acts while two surfaces slide a distance $d$ past each other, it turns mechanical energy into thermal energy: $\Delta E_{th} = f_k d$. Then $K_i + U_i = K_f + U_f + \Delta E_{th}$.
Why it works
Friction comes from countless tiny bumps snagging and letting go, which sets the atoms near both surfaces jiggling faster. Faster random jiggling is higher temperature. The energy isn't destroyed; it changes from the ordered motion of the whole box into the disordered motion of its atoms and the floor's. That's why rubbing your hands together warms them.
Worked examples
Worked example: Thermal energy produced
A $15\ \text{kg}$ box is dragged $8\ \text{m}$ across a floor with $\mu_k = 0.3$. How much thermal energy is produced?
- $f_k = \mu_k mg = 0.3 \times 147 = 44.1\ \text{N}$.
- $\Delta E_{th} = f_k d = 44.1 \times 8 \approx 353\ \text{J}$.
Answer: about $353\ \text{J}$
Worked example: Sliding to a stop
A puck slides at $5\ \text{m/s}$ onto rough ice with $\mu_k = 0.1$. How far does it go?
- All its kinetic energy becomes thermal energy: $\tfrac12 mv^2 = \mu_k mg\,d$.
- $d = \dfrac{v^2}{2\mu_k g} = \dfrac{25}{1.96} \approx 12.8\ \text{m}$.
Answer: about $12.8\ \text{m}$
Worked example: A rough slope
A $20\ \text{kg}$ sled starts from rest at the top of a slope $6\ \text{m}$ high and slides $15\ \text{m}$ down it against $30\ \text{N}$ of friction. Find its speed at the bottom.
- Start: $U_g = (20)(9.8)(6) = 1{,}176\ \text{J}$.
- Thermal energy: $30 \times 15 = 450\ \text{J}$.
- $K = 1176 - 450 = 726\ \text{J}$, so $v = \sqrt{\dfrac{2(726)}{20}} \approx 8.52\ \text{m/s}$.
Answer: about $8.52\ \text{m/s}$
Common mistake
Thinking energy disappears when friction stops a sliding box. It is still there, as thermal energy in the box and the floor. Check: the starting mechanical energy should equal the final mechanical energy plus $f_k d$; if your bookkeeping is short, you left out the thermal energy.
The Science Lab's Energy and the First Law page follows this energy further, into heat and the first law of thermodynamics. It writes the first law as $\Delta U = Q - W$, with $W$ the work done by a gas; Physics 2 writes $\Delta U = Q + W$, with $W$ the work done on the gas. They say the same thing with opposite signs for $W$.