A ship of mass `3 times 10^7 kg` initially at rest is pulled by a force of `5 times 10^4 N` through a distance of 3m. Assume that the resistance due to water is negligible, the speed of the ship is

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Mass of the ship (M) = \(3 \times 10^7 \, \text{kg}\)
- Force applied (F) = \(5 \times 10^4 \, \text{N}\)
- Initial velocity (U) = \(0 \, \text{m/s}\) (the ship is initially at rest)
- Displacement (S) = \(3 \, \text{m}\)

### Step 2: Calculate the acceleration of the ship
Using Newton's second law of motion, we know that:
\[ F = M \cdot a \]
Where \(a\) is the acceleration. Rearranging this gives us:
\[ a = \frac{F}{M} \]
Substituting the values:
\[ a = \frac{5 \times 10^4 \, \text{N}}{3 \times 10^7 \, \text{kg}} = \frac{5}{3} \times 10^{-3} \, \text{m/s}^2 \]

### Step 3: Use the kinematic equation to find the final velocity
We will use the kinematic equation:
\[ V^2 = U^2 + 2AS \]
Since the initial velocity \(U = 0\), this simplifies to:
\[ V^2 = 2AS \]
Substituting the values of \(A\) and \(S\):
\[ V^2 = 2 \left(\frac{5}{3} \times 10^{-3}\right) \cdot 3 \]
Calculating this gives:
\[ V^2 = 2 \cdot \frac{5}{3} \cdot 3 \times 10^{-3} = 10 \times 10^{-3} = 10^{-2} \]

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

### Final Answer
The speed of the ship is \(0.1 \, \text{m/s}\).