NLM BYU 33

1. System Setup and Coordinate Convention

Let us define the coordinate system with the East direction as positive ($+\hat{i}$) and West as negative ($-\hat{i}$).

• Mass: $m = 5.0 \, \text{kg}$
• Initial Velocity: $\vec{u} = +8.0 \, \hat{i} \, \text{m/s}$
• External Force: $\vec{F} = -kt \, \hat{i}$, where $k = 5.0 \, \text{N/s}$. So, $\vec{F} = -5t \, \hat{i}$.
• Coefficient of Friction: $\mu = 0.3$
• Gravity: $g = 10 \, \text{m/s}^2$

First, we calculate the normal force ($N$) and the limiting friction ($f_{\text{max}}$):

$$ N = mg = 5.0 \times 10 = 50 \, \text{N} $$ $$ f_{\text{max}} = \mu N = 0.3 \times 50 = 15 \, \text{N} $$

2. Phase 1: Deceleration to Rest ($0 \le t \le t_0$)

Initially, the block moves East ($v > 0$). Therefore, kinetic friction acts West ($-\hat{i}$). The external force is also acting West.

$$ \vec{f}_k = -15 \, \hat{i} \, \text{N} $$ $$ \vec{F}_{\text{net}} = \vec{F} + \vec{f}_k = (-5t – 15) \, \hat{i} $$

Using Newton’s Second Law ($\vec{F}_{\text{net}} = m\vec{a}$):

$$ \vec{a} = \frac{-5t – 15}{5} \, \hat{i} = -(t + 3) \, \hat{i} \, \text{m/s}^2 $$

We integrate the acceleration to find the velocity as a function of time:

$$ v(t) = u + \int_{0}^{t} a(t’) \, dt’ $$ $$ v(t) = 8 – \int_{0}^{t} (t’ + 3) \, dt’ = 8 – \left[ \frac{t’^2}{2} + 3t’ \right]_0^t $$ $$ v(t) = 8 – \frac{t^2}{2} – 3t $$

To find the time $t_0$ when the block stops momentarily, set $v(t) = 0$:

$$ \frac{t^2}{2} + 3t – 8 = 0 \implies t^2 + 6t – 16 = 0 $$ $$ (t + 8)(t – 2) = 0 $$

Since time must be positive, the block stops at $t_0 = 2.0 \, \text{s}$.

3. Phase 2: Static Region ($2.0 \, \text{s} \le t \le t_{\text{slip}}$)

At $t = 2.0 \, \text{s}$, the block comes to rest ($v=0$). We must check if the external force is sufficient to overcome static friction immediately.

• External driving force at $t=2$: $\vec{F}_{\text{ext}} = -5(2) \, \hat{i} = -10 \, \hat{i} \, \text{N}$ (Westward push of 10 N).
• Maximum Static Friction: $f_s^{\text{max}} = 15 \, \text{N}$.

Since the driving force magnitude ($10 \, \text{N}$) is less than the maximum static friction ($15 \, \text{N}$), the block remains at rest.

For the block to remain stationary ($\vec{a}=0$), the static friction must exactly balance the external force. Since $\vec{F}_{\text{ext}}$ is West, friction must act East (positive).

$$ \vec{f}_s + \vec{F}_{\text{ext}} = 0 \implies \vec{f}_s = -\vec{F}_{\text{ext}} = -(-5t) \, \hat{i} = +5t \, \hat{i} $$

This static phase continues until the required friction equals the maximum limit:

$$ f_s = 5t = 15 \implies t = 3.0 \, \text{s} $$

4. Phase 3: Motion Resumes Westward ($t > 3.0 \, \text{s}$)

For $t > 3.0 \, \text{s}$, the external driving force magnitude $|F| = 5t > 15 \, \text{N}$. The force overcomes friction, and the block accelerates Westwards ($v < 0$).

Since the relative motion is Westward, kinetic friction acts Eastward ($+\hat{i}$) with a constant magnitude.

5. Graphical Representation

The variation of frictional force $f$ with time $t$ is summarized as:

$$ f(t) = \begin{cases} -15 \, \text{N} & 0 \le t < 2 \, \text{s} \quad (\text{Kinetic, West}) \\ +5t \, \text{N} & 2 \le t \le 3 \, \text{s} \quad (\text{Static, East}) \\ +15 \, \text{N} & t > 3 \, \text{s} \quad (\text{Kinetic, East}) \end{cases} $$