Solution to Question 17
1. Analyzing Friction Forces
The problem states distinct coefficients depending on the sliding direction:
- Sliding Right: When $v_{block} > v_{belt}$, the block slides right relative to the belt. Friction acts to the Left. Coefficient $\mu_r$. Deceleration magnitude $a_r = \mu_r g$.
- Sliding Left: When $v_{block} < v_{belt}$, the block slides left relative to the belt. Friction acts to the Right. Coefficient $\mu_l$. Acceleration magnitude $a_l = \mu_l g$.
Given: $\mu_r = 0.25, \mu_l = 0.50$. Thus, acceleration to the right ($a_l$) is twice as strong as deceleration to the left ($a_r$).
2. Cycle Analysis
In the steady state, the block’s velocity profile repeats every $2\tau$. Due to the large coefficients and sufficient time $\tau=1.0\text{s}$, the block reaches the belt velocity and sticks during each half-cycle.
Phase 1 (Belt velocity $+v_0$):
The block starts from $-v_0$ (end of previous cycle). It accelerates right with $a_l = \mu_l g$.
Time to reach $v_0$: $t_1 = \frac{v_0 – (-v_0)}{a_l} = \frac{2v_0}{a_l}$.
For the remainder of the time $(\tau – t_1)$, it travels at $v_0$.
Phase 2 (Belt velocity $-v_0$):
The block starts from $v_0$. It accelerates left (decelerates) with magnitude $a_r = \mu_r g$.
Time to reach $-v_0$: $t_2 = \frac{v_0 – (-v_0)}{a_r} = \frac{2v_0}{a_r}$.
For the remainder of the time $(\tau – t_2)$, it travels at $-v_0$.
3. Calculating Displacement and Average Velocity
The average velocity is $\bar{v} = \frac{\text{Net Displacement}}{\text{Total Time}}$. The total time is $2\tau$.
Displacement in Phase 1 ($X_1$):
The velocity changes linearly from $-v_0$ to $v_0$ (symmetric about 0, so net displacement during acceleration is 0), then stays at $v_0$.
$$ X_1 = 0 + v_0 (\tau – t_1) = v_0 \left(\tau – \frac{2v_0}{a_l}\right) $$
Displacement in Phase 2 ($X_2$):
The velocity changes linearly from $v_0$ to $-v_0$ (symmetric about 0, displacement 0), then stays at $-v_0$.
$$ X_2 = 0 + (-v_0) (\tau – t_2) = -v_0 \left(\tau – \frac{2v_0}{a_r}\right) $$
Net Displacement ($X_{net}$):
$$ X_{net} = X_1 + X_2 = v_0 \tau – \frac{2v_0^2}{a_l} – v_0 \tau + \frac{2v_0^2}{a_r} = 2v_0^2 \left( \frac{1}{a_r} – \frac{1}{a_l} \right) $$
Average Velocity:
$$ \bar{v} = \frac{X_{net}}{2\tau} = \frac{v_0^2}{\tau} \left( \frac{1}{a_r} – \frac{1}{a_l} \right) $$
Substituting $a_r = \mu_r g$ and $a_l = \mu_l g$:
$$ \bar{v} = \frac{v_0^2}{\tau g} \left( \frac{1}{\mu_r} – \frac{1}{\mu_l} \right) $$
4. Numerical Calculation
Given: $v_0 = 1.0$, $\tau = 1.0$, $g = 10$, $\mu_r = 0.25$, $\mu_l = 0.50$.
$$ \bar{v} = \frac{1.0^2}{1.0 \times 10} \left( \frac{1}{0.25} – \frac{1}{0.50} \right) $$
$$ \bar{v} = 0.1 \times (4 – 2) = 0.1 \times 2 = 0.2 \text{ m/s} $$
Answer: $$ 0.2 \text{ m/s} $$
The problem states distinct coefficients depending on the sliding direction:
- Sliding Right: When $v_{block} > v_{belt}$, the block slides right relative to the belt. Friction acts to the Left. Coefficient $\mu_r$. Deceleration magnitude $a_r = \mu_r g$.
- Sliding Left: When $v_{block} < v_{belt}$, the block slides left relative to the belt. Friction acts to the Right. Coefficient $\mu_l$. Acceleration magnitude $a_l = \mu_l g$.
In the steady state, the block’s velocity profile repeats every $2\tau$. Due to the large coefficients and sufficient time $\tau=1.0\text{s}$, the block reaches the belt velocity and sticks during each half-cycle.
Phase 1 (Belt velocity $+v_0$):
The block starts from $-v_0$ (end of previous cycle). It accelerates right with $a_l = \mu_l g$.
Time to reach $v_0$: $t_1 = \frac{v_0 – (-v_0)}{a_l} = \frac{2v_0}{a_l}$.
For the remainder of the time $(\tau – t_1)$, it travels at $v_0$.
Phase 2 (Belt velocity $-v_0$):
The block starts from $v_0$. It accelerates left (decelerates) with magnitude $a_r = \mu_r g$.
Time to reach $-v_0$: $t_2 = \frac{v_0 – (-v_0)}{a_r} = \frac{2v_0}{a_r}$.
For the remainder of the time $(\tau – t_2)$, it travels at $-v_0$.
The average velocity is $\bar{v} = \frac{\text{Net Displacement}}{\text{Total Time}}$. The total time is $2\tau$.
Displacement in Phase 1 ($X_1$): The velocity changes linearly from $-v_0$ to $v_0$ (symmetric about 0, so net displacement during acceleration is 0), then stays at $v_0$. $$ X_1 = 0 + v_0 (\tau – t_1) = v_0 \left(\tau – \frac{2v_0}{a_l}\right) $$
Displacement in Phase 2 ($X_2$): The velocity changes linearly from $v_0$ to $-v_0$ (symmetric about 0, displacement 0), then stays at $-v_0$. $$ X_2 = 0 + (-v_0) (\tau – t_2) = -v_0 \left(\tau – \frac{2v_0}{a_r}\right) $$
Net Displacement ($X_{net}$): $$ X_{net} = X_1 + X_2 = v_0 \tau – \frac{2v_0^2}{a_l} – v_0 \tau + \frac{2v_0^2}{a_r} = 2v_0^2 \left( \frac{1}{a_r} – \frac{1}{a_l} \right) $$
Average Velocity: $$ \bar{v} = \frac{X_{net}}{2\tau} = \frac{v_0^2}{\tau} \left( \frac{1}{a_r} – \frac{1}{a_l} \right) $$ Substituting $a_r = \mu_r g$ and $a_l = \mu_l g$: $$ \bar{v} = \frac{v_0^2}{\tau g} \left( \frac{1}{\mu_r} – \frac{1}{\mu_l} \right) $$
Given: $v_0 = 1.0$, $\tau = 1.0$, $g = 10$, $\mu_r = 0.25$, $\mu_l = 0.50$. $$ \bar{v} = \frac{1.0^2}{1.0 \times 10} \left( \frac{1}{0.25} – \frac{1}{0.50} \right) $$ $$ \bar{v} = 0.1 \times (4 – 2) = 0.1 \times 2 = 0.2 \text{ m/s} $$