Problem — Work against friction

Work against friction. A $20\;\mathrm{kg}$ box is dragged across a horizontal floor over $10\;\mathrm{m}$, with $\mu_d = 0{,}3$. Calculate the work done against friction. How much of that energy is recoverable as motion?

Solution

Data: $m = 20\;\mathrm{kg}$, $s = 10\;\mathrm{m}$, $\mu_d = 0{,}3$. On a horizontal surface the normal reaction equals the weight: $N = mg$.

Kinetic friction force:

$f = \mu_d N = \mu_d\,mg = 0{,}3\cdot 20\cdot 9{,}81 = 58{,}9\;\mathrm{N}$

Work done against friction (to overcome it, the applied force travels $s$ opposing $f$):

$W = f\,s = 58{,}9\cdot 10 \approx 589\;\mathrm{J}$

$\ev{W \approx 589\;\mathrm{J}}$

Energy recoverable as motion:

$\ev{0\;\mathrm{J}}$

Physical reading. Kinetic friction is a dissipative, non-conservative force: all $589\;\mathrm{J}$ turn into heat that warms the box and the floor and disperses into the environment. None of it is recoverable as ordered kinetic energy: unlike a spring, friction has no reservoir from which to return energy. It is the thermal price of displacement.

Collegamenti

Argomenti: Work and energy Concetti: Thermal energy from friction · Work of a force · Kinetic friction Oggetti: Block