A mass of 1 kg is acted upon by a single force `F=(4hati+4hatj)N`. Under this force it is displaced from (0,0) to (1m,1m). If initially the speed of the particle was 2 `ms^(-1)`, its final speed should be

Options

A. `6 ms^(-1)`
B. `4.5 ms^(-1)`
C. `8 ms^(-1)`
D. `4 ms^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Mass (m) = 1 kg
- Force (F) = \(4 \hat{i} + 4 \hat{j}\) N
- Initial speed (V_i) = 2 m/s
- Initial position = (0,0)
- Final position = (1,1)

### Step 2: Calculate the displacement vector
The displacement vector \( \vec{S} \) can be calculated as:
\[ \vec{S} = \text{Final position} - \text{Initial position} = (1, 1) - (0, 0) = 1 \hat{i} + 1 \hat{j} \]

### Step 3: Calculate the work done by the force
The work done \( W \) by the force is given by the dot product of the force and the displacement:
\[ W = \vec{F} \cdot \vec{S} = (4 \hat{i} + 4 \hat{j}) \cdot (1 \hat{i} + 1 \hat{j}) \]
Calculating the dot product:
\[ W = 4 \cdot 1 + 4 \cdot 1 = 4 + 4 = 8 \text{ J} \]

### Step 4: Use the work-energy theorem
According to the work-energy theorem:
\[ W = \Delta KE = KE_f - KE_i \]
Where:
- \( KE_f = \frac{1}{2} m V_f^2 \) (final kinetic energy)
- \( KE_i = \frac{1}{2} m V_i^2 \) (initial kinetic energy)

Substituting the values:
\[ 8 = \frac{1}{2} \cdot 1 \cdot V_f^2 - \frac{1}{2} \cdot 1 \cdot (2)^2 \]
This simplifies to:
\[ 8 = \frac{1}{2} V_f^2 - 2 \]

### Step 5: Solve for \( V_f^2 \)
Rearranging the equation:
\[ 8 + 2 = \frac{1}{2} V_f^2 \]
\[ 10 = \frac{1}{2} V_f^2 \]
Multiplying both sides by 2:
\[ 20 = V_f^2 \]

### Step 6: Find \( V_f \)
Taking the square root:
\[ V_f = \sqrt{20} \approx 4.47 \text{ m/s} \]

### Step 7: Round to the nearest option
Since the closest option provided is 4.5 m/s, we conclude:
\[ \text{Final speed } V_f \approx 4.5 \text{ m/s} \]

### Final Answer
The final speed of the particle is approximately **4.5 m/s**.

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