Solution to Question 32
First, consider the forces acting on the block. The vertical forces are weight $mg$ and normal reaction $N$. Since the floor is horizontal, $N = mg$. The friction force is $f = \mu N = \mu mg$. By Newton’s second law: $$F_{net} = ma$$ $$-\mu mg = ma \implies a = -\mu g$$ Interestingly, the acceleration (retardation) is independent of the mass. Even though the mass $m$ is decreasing, both the friction force and the inertia decrease proportionally. Therefore, the block slows down at a constant rate $a = -\mu g$.
The motion of the block will end when either of two events happens first:
- Kinematic Stop: The velocity reaches zero due to friction.
- Material Vanished: The mass reduces to zero due to wear.
Distance to Kinematic Stop ($x_1$):
Using $v^2 – u^2 = 2ax$ with $v=0$ and $a=-\mu g$:
$$0 – u^2 = 2(-\mu g) x_1$$
$$x_1 = \frac{u^2}{2\mu g}$$
Distance to Material Vanish ($x_2$):
The mass decreases at a rate $r$ per unit distance: $\frac{dm}{dx} = -r$.
The mass at distance $x$ is $m(x) = m_0 – rx$.
The block vanishes when $m(x) = 0$:
$$m_0 – r x_2 = 0 \implies x_2 = \frac{m_0}{r}$$
The total distance travelled $x$ is determined by whichever limit is reached first.
$$x = \min\left( \frac{m}{r}, \frac{u^2}{2\mu g} \right)$$Answer: $x = \frac{m}{r}$ or $x = \frac{u^2}{2\mu g}$, whichever is smaller.