If a ship of mass ` 4 xx 10 ^( 7 ) ` kg initially at rest is pulled by a force of ` 5 xx 10 ^ ( 4 ) ` N through a distance of 4 m, then the speed of the ship will be (resistance due to water is negligible )

Options

A. `1.5 ms^(-1)`
B. `60 ms^(-1)`
C. `0.1 ms^(-1)`
D. `5 ms^(-1)`

Explanation

To solve the problem step by step, we will use the concepts of Newton's second law of motion and the equations of motion.

### Step 1: Identify the given values
- Mass of the ship, \( m = 4 \times 10^7 \) kg
- Force applied, \( F = 5 \times 10^4 \) N
- Initial velocity, \( u = 0 \) m/s (the ship is initially at rest)
- Distance moved, \( s = 4 \) m

### Step 2: Calculate the acceleration
Using Newton's second law, the acceleration \( a \) can be calculated using the formula:
\[ a = \frac{F}{m} \]
Substituting the known values:
\[ a = \frac{5 \times 10^4}{4 \times 10^7} \]
Calculating this gives:
\[ a = \frac{5}{4} \times 10^{-3} \text{ m/s}^2 = 1.25 \times 10^{-3} \text{ m/s}^2 \]

### Step 3: Use the equation of motion to find the final velocity
We will use the equation of motion:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ v^2 = 0^2 + 2 \times (1.25 \times 10^{-3}) \times 4 \]
Calculating this gives:
\[ v^2 = 0 + 2 \times 1.25 \times 10^{-3} \times 4 \]
\[ v^2 = 10 \times 10^{-3} = 1 \times 10^{-2} \]

### Step 4: Calculate the final velocity
Taking the square root of both sides:
\[ v = \sqrt{1 \times 10^{-2}} = 0.1 \text{ m/s} \]

### Final Answer
The speed of the ship after being pulled through a distance of 4 m is \( 0.1 \) m/s.

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