Solution: Dynamics of a Box with Variable Force

1. Physics Principles

This problem involves Newton’s Laws of Motion with time-varying forces and friction.

• Friction: Static friction prevents motion until the applied force exceeds the limiting friction ($f_L = \mu_s N$). Kinetic friction ($f_k = \mu_k N$) opposes motion once it starts.
• Impulse-Momentum: The change in velocity is determined by the net impulse divided by mass ($\Delta v = \frac{1}{m} \int F_{net} dt$).
• Work-Energy & Power: Power is the rate of doing work. Average power is total work divided by the time interval.

2. Visual Analysis

3. Step-by-Step Calculation

Step 1: Analyze Friction

Given: $m = 4$ kg, $\mu_s = \mu_k = 0.25$.

$$ f_{limit} = f_k = \mu m g = 0.25 \times 4 \times 10 = 10 \text{ N} $$

The box will not move until the applied force $F > 10$ N.

Step 2: Analyze Option (a) – Start of Motion

From the graph, $F(t)$ increases linearly from 0 to 20 N over 4 seconds. The equation for the first 4 seconds is:

$$ F(t) = \frac{20}{4}t = 5t $$

Motion starts when $F(t) = 10$ N:

$$ 5t = 10 \implies t = 2.0 \text{ s} $$

Conclusion (a) is correct.

Step 3: Analyze Option (b) – Maximum Velocity

The box accelerates while $F(t) > f_k$. This occurs between $t = 2$ s and the time when $F$ drops back to 10 N.

From $t=4$ to $t=8$, the equation is $F(t) = 20 – 5(t-4)$. Setting $F=10$:

$$ 20 – 5(t-4) = 10 \implies 5(t-4) = 10 \implies t = 6 \text{ s} $$

Velocity is maximum at $t=6$ s (when acceleration becomes zero).

To find $v_{max}$, we use the impulse-momentum theorem ($m\Delta v = \int F_{net} dt$). The net impulse is the area under the F-t graph minus the impulse of friction, from $t=2$ to $t=6$.

• Area of F-t (Trapezoid 2s to 6s): Area = Area(2-4) + Area(4-6) = $30 + 30 = 60$ Ns.
• Impulse of Friction: $f_k \times \Delta t = 10 \times (6 – 2) = 40$ Ns.

$$ m v_{max} = 60 – 40 = 20 \text{ Ns} $$ $$ 4 \times v_{max} = 20 \implies v_{max} = 5.0 \text{ m/s} $$

Conclusion (b) is correct.

Step 4: Analyze Option (c) – Stopping Time

After $t=6$ s, the net force is retarding ($F < 10$). The box stops when the total change in momentum cancels the gained momentum. We need a retarding impulse of -20 Ns.

Let’s check the interval $t=6$ to $t=8$:

• Area of F (Triangle base 2, height 10): $\frac{1}{2} \times 2 \times 10 = 10$ Ns.
• Friction Impulse: $10 \times 2 = 20$ Ns.
• Net Impulse: $10 – 20 = -10$ Ns.

At $t=8$ s, velocity corresponds to $20 – 10 = 10$ Ns momentum remaining. $v = 2.5$ m/s.

After $t=8$, $F=0$. Only friction acts ($a = -10/4 = -2.5 \text{ m/s}^2$).

$$ v = u + at \implies 0 = 2.5 – 2.5(t_{stop} – 8) $$ $$ t_{stop} – 8 = 1 \implies t_{stop} = 9.0 \text{ s} $$

Conclusion (c) is correct.

Step 5: Analyze Option (d) – Average Power

The box moves from $t=2$ to $t=9$. Total kinetic energy change is zero, so the total work done by Force $F$ plus Work done by Friction is zero: $W_F + W_f = 0 \implies W_F = |W_f|$.

Average Power is Total Work / Time Duration.

• For Force F: It does work from $t=2$ to $t=8$ (while box is moving and F is non-zero). $\Delta t_F = 6$ s.
• For Friction f: It does work from $t=2$ to $t=9$ (entire motion). $\Delta t_f = 7$ s.

$$ P_{avg, F} = \frac{W_F}{6} \quad \text{and} \quad P_{avg, f} = \frac{|W_f|}{7} $$

Since $W_F = |W_f|$, clearly $P_{avg, F} > P_{avg, f}$.

Conclusion (d) is correct.