Solution: Disc Sliding with Variable Friction
1. Work-Energy Theorem
The disc starts with speed \( u \) and stops after distance \( S \). The change in kinetic energy is equal to the work done by the friction force.
$$ \Delta K = W_{friction} $$ $$ 0 – \frac{1}{2}mu^2 = -\int_{0}^{S} f_r dx $$
2. Integrating Friction
The friction force is \( f_r = \mu N = (\mu_0 + kx)mg \).
$$ \frac{1}{2}mu^2 = \int_{0}^{S} (\mu_0 + kx)mg \, dx $$ $$ \frac{u^2}{2g} = \left[ \mu_0 x + \frac{1}{2}kx^2 \right]_0^S $$ $$ \frac{u^2}{2g} = \mu_0 S + \frac{1}{2}k S^2 $$
3. Solving for Distance S
Rearranging into a standard quadratic equation for \( S \):
$$ k S^2 + 2\mu_0 S – \frac{u^2}{g} = 0 $$Using the quadratic formula \( S = \frac{-b + \sqrt{b^2 – 4ac}}{2a} \):
$$ S = \frac{-2\mu_0 + \sqrt{4\mu_0^2 – 4k(-u^2/g)}}{2k} $$ $$ S = \frac{-2\mu_0 + 2\sqrt{\mu_0^2 + \frac{ku^2}{g}}}{2k} $$
$$ S = \frac{\sqrt{k u^2 g + (\mu_0 g)^2} – \mu_0 g}{kg} $$