A train of mass 100 metric tons is drawn up an incline of 1 in 49 at the rate of 36 kph by a engine. If the resistance due to friction be 10N per metric ton, calculate the power of the engine if the steam is shut off, how far will the train move before it comes to rest ?

Answer

To solve the problem step by step, we will break it down into manageable parts.

### Step 1: Convert the speed from km/h to m/s
The speed of the train is given as 36 km/h. We need to convert this to meters per second (m/s).

\[ \text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18} \]
\[ \text{Speed in m/s} = 36 \times \frac{5}{18} = 10 \text{ m/s} \]

**Hint:** Remember that to convert km/h to m/s, you multiply by \(\frac{5}{18}\).

### Step 2: Calculate the total resistance acting on the train
The resistance due to friction is given as 10 N per metric ton. Since the mass of the train is 100 metric tons, the total frictional force can be calculated as follows:

\[ \text{Frictional force} = \text{mass} \times \text{friction per metric ton} = 100 \times 10 = 1000 \text{ N} \]

**Hint:** To find the total frictional force, multiply the mass of the train by the frictional force per unit mass.

### Step 3: Calculate the component of gravitational force acting down the incline
The incline is given as 1 in 49. This means for every 49 units of horizontal distance, there is a rise of 1 unit. The sine of the angle of incline can be calculated as:

\[ \sin \theta = \frac{1}{49} \]

The component of gravitational force acting down the incline can be calculated using:

\[ \text{Weight component down the incline} = \text{mass} \times g \times \sin \theta \]
Assuming \(g = 9.81 \, \text{m/s}^2\):

\[ \text{Weight component} = 100 \times 9.81 \times \frac{1}{49} \approx 20 \text{ N} \]

**Hint:** Use the sine of the angle to find the component of gravitational force acting along the incline.

### Step 4: Calculate the total deceleration
The total deceleration \(a\) acting on the train is the sum of the deceleration due to friction and the deceleration due to gravity:

\[ \text{Total deceleration} = \text{frictional force} + \text{weight component} \]
\[ a = 1000 + 20 = 1020 \text{ N} \]

Since the mass of the train is 100 metric tons (or 100,000 kg), we convert the total deceleration to acceleration:

\[ \text{Total deceleration (in m/s}^2\text{)} = \frac{1020}{100} = 10.2 \text{ m/s}^2 \]

**Hint:** Remember that deceleration is the total force divided by mass.

### Step 5: Use the kinematic equation to find the distance traveled before coming to rest
We can use the equation:

\[ v^2 = u^2 + 2as \]

Where:
- \(v = 0\) (final velocity)
- \(u = 10 \, \text{m/s}\) (initial velocity)
- \(a = -10.2 \, \text{m/s}^2\) (deceleration, negative because it opposes motion)

Rearranging gives:

\[ 0 = (10)^2 + 2(-10.2)s \]
\[ 0 = 100 - 20.4s \]
\[ 20.4s = 100 \]
\[ s = \frac{100}{20.4} \approx 4.90 \text{ m} \]

**Hint:** When using the kinematic equation, remember to account for the signs of velocity and acceleration correctly.

### Final Answer
The train will move approximately **4.90 meters** before coming to rest.