1. System Analysis and Force Diagram

We analyze the motion of two bodies: the barge (mass $M = 200 \text{ kg}$) and the block (mass $m = 50 \text{ kg}$). The motor pulls the block with a constant force through an inextensible string.

• Forces on the Block:
  • Tension ($T$) acting towards the motor (Left). Given $T = 150 \text{ N}$.
  • Friction ($f$) opposes motion relative to the deck. Since the block moves towards the motor, friction acts to the Right.
  • Normal force $N = mg = 50 \times 10 = 500 \text{ N}$.
  • Kinetic friction $f = \mu N = 0.2 \times 500 = 100 \text{ N}$.
• Forces on the Barge:
  • By Newton’s 3rd Law, the string exerts a force $T$ on the motor (attached to the barge) towards the right.
  • The friction force exerted by the block on the barge is equal and opposite to the friction on the block. So, friction on the barge acts to the Left ($f = 100 \text{ N}$).
  • Water offers no resistance.

2. Acceleration Calculation (0 to 4 s)

For the Barge ($M=200 \text{ kg}$):
Net force $F_{net,B} = T – f = 150 – 100 = 50 \text{ N}$ (Right).
Acceleration $a_B = \frac{F_{net,B}}{M} = \frac{50}{200} = 0.25 \text{ m/s}^2$.

For the Block ($m=50 \text{ kg}$):
Net force magnitude $F_{net,b} = T – f = 150 – 100 = 50 \text{ N}$ (Left).
Acceleration $a_b = \frac{50}{50} = 1 \text{ m/s}^2$ (Left).

Relative Acceleration:
$a_{rel} = a_B – (-a_b) = 0.25 – (-1) = 1.25 \text{ m/s}^2$.

3. Motion up to t = 4 s

Since the motor is on for 4 seconds:

• Velocity of Barge: $v_B(4) = u + a_B t = 0 + 0.25(4) = 1 \text{ m/s}$.
• Velocity of Block: $v_b(4) = 0 + 1(4) = 4 \text{ m/s}$ (Left).
• Relative Velocity: $v_{rel}(4) = v_B – v_{b,vector} = 1 – (-4) = 5 \text{ m/s}$.
• Displacement of Barge: $S_1 = \frac{1}{2} a_B t^2 = \frac{1}{2}(0.25)(16) = 2 \text{ m}$.
• Relative Displacement: $S_{rel,1} = \frac{1}{2} a_{rel} t^2 = \frac{1}{2}(1.25)(16) = 10 \text{ m}$.

4. Motion after t = 4 s (Motor Off)

When the motor is switched off, Tension $T = 0$. Friction continues to act as long as there is relative motion.

• Deceleration of Barge: Force is now only friction to the Left ($f=100 \text{ N}$).
  $a’_B = \frac{-100}{200} = -0.5 \text{ m/s}^2$.
• Deceleration of Block: Force is only friction to the Right ($f=100 \text{ N}$).
  $a’_b = \frac{100}{50} = 2 \text{ m/s}^2$ (Right).
• Relative Deceleration: Relative acceleration opposing motion: $a_{stop} = |a’_B| + |a’_b| = 0.5 + 2 = 2.5 \text{ m/s}^2$.

Time to stop relative motion:
$v_{final} = v_{initial} – a_{stop} t’$
$0 = 5 – 2.5 t’ \Rightarrow t’ = 2 \text{ s}$.
Total time = $4 + 2 = 6 \text{ s}$.

During this interval ($t=4$ to $6$ s), the barge moves:
$S_2 = v_B(4)t’ + \frac{1}{2} a’_B (t’)^2 = 1(2) + \frac{1}{2}(-0.5)(2^2) = 2 – 1 = 1 \text{ m}$.

Relative displacement during this interval:
$S_{rel,2} = \frac{v_{rel}^2}{2 a_{stop}} = \frac{5^2}{2(2.5)} = \frac{25}{5} = 5 \text{ m}$.

5. Final Results

Total Displacement on the Deck (Relative Displacement):
$S_{total\_rel} = S_{rel,1} + S_{rel,2} = 10 + 5 = 15 \text{ m}$.

Total Displacement of the Barge (w.r.t Water):
$S_{barge} = S_1 + S_2 = 2 + 1 = 3 \text{ m}$.

6. Velocity-Time Graph for the Barge

The barge accelerates uniformly from 0 to 1 m/s in 4 seconds, then decelerates uniformly to 0 m/s in 2 seconds.