A ship of mass `3xx10^(2)kg` initially at rest is pulled by a force of `5xx10^(4)` N through a distance of 3m. Neglecting frcition, the speed of the ship at this moment is:

Options

A. `31.62m//s`
B. `1.55m//s`
C. `0.15m//s`
D. `2m//s`

Explanation

To solve the problem, we will use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Here are the steps to find the speed of the ship:

### Step 1: Identify the known values
- Mass of the ship, \( m = 3 \times 10^2 \, \text{kg} = 300 \, \text{kg} \)
- Force applied, \( F = 5 \times 10^4 \, \text{N} \)
- Distance moved, \( d = 3 \, \text{m} \)
- Initial speed, \( u = 0 \, \text{m/s} \) (the ship is initially at rest)

### Step 2: Calculate the work done by the force
The work done \( W \) by the force can be calculated using the formula:
\[ W = F \times d \]
Substituting the known values:
\[ W = (5 \times 10^4 \, \text{N}) \times (3 \, \text{m}) = 1.5 \times 10^5 \, \text{J} \]

### Step 3: Apply the work-energy theorem
According to the work-energy theorem:
\[ W = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]
Since the ship starts from rest, the initial kinetic energy \( KE_{\text{initial}} = 0 \):
\[ W = KE_{\text{final}} - 0 = KE_{\text{final}} \]
Thus, we have:
\[ 1.5 \times 10^5 \, \text{J} = KE_{\text{final}} \]

### Step 4: Express kinetic energy in terms of speed
The kinetic energy \( KE \) is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
Substituting for \( KE_{\text{final}} \):
\[ 1.5 \times 10^5 \, \text{J} = \frac{1}{2} (300 \, \text{kg}) v^2 \]

### Step 5: Solve for \( v^2 \)
Rearranging the equation to solve for \( v^2 \):
\[ 1.5 \times 10^5 = 150 v^2 \]
Dividing both sides by 150:
\[ v^2 = \frac{1.5 \times 10^5}{150} = 1000 \]

### Step 6: Calculate the final speed \( v \)
Taking the square root of both sides:
\[ v = \sqrt{1000} \approx 31.62 \, \text{m/s} \]

### Final Answer
The speed of the ship at this moment is approximately \( 31.62 \, \text{m/s} \).

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