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Q7(b):
A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (b) the acceleration of the train.

Solution :

Step 1: Identification of Given Variables

To determine the acceleration of the train, we define the parameters provided in the problem statement:

  • Mass of the engine ($m_e$) = $8000\text{ kg}$
  • Number of wagons ($n$) = $5$
  • Mass per wagon ($m_w$) = $2000\text{ kg}$
  • Applied force by engine ($F_{applied}$) = $40000\text{ N}$
  • Friction force ($F_{friction}$) = $5000\text{ N}$

Step 2: Calculating Total Mass of the System

The system consists of the engine and the five wagons moving together as a single unit. Therefore, the total mass ($M_{total}$) is the sum of the mass of the engine and the total mass of all wagons.

$M_{total} = m_e + (n \times m_w)$

$M_{total} = 8000\text{ kg} + (5 \times 2000\text{ kg})$

$M_{total} = 8000\text{ kg} + 10000\text{ kg} = 18000\text{ kg}$

Step 3: Calculating Net Force

According to [Newton’s Second Law of Motion], the net force ($F_{net}$) acting on an object is the vector sum of all forces applied to it. Since friction acts in the direction opposite to the applied force:

$F_{net} = F_{applied} - F_{friction}$

$F_{net} = 40000\text{ N} - 5000\text{ N} = 35000\text{ N}$

Step 4: Calculation of Acceleration

Applying the formula derived from [Newton’s Second Law]: $F_{net} = M_{total} \times a$, we rearrange to solve for acceleration ($a$):

$a = \frac{F_{net}}{M_{total}}$

$a = \frac{35000\text{ N}}{18000\text{ kg}}$

$a = \frac{35}{18}\text{ m/s}^2 \approx 1.944\text{ m/s}^2$

Visual Representation of Force Vectors

Train System (18000 kg) F_applied = 40000 N F_friction = 5000 N

Final Solution: The acceleration of the train is $1.944\text{ m/s}^2$.


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