A particle of mass 0.01 kg travels along a space curve with velocity given by `4hat(i)+16 hat (k)m//s` After some time its velocity becomes `8hat(i)+20hat(j)m//s` due to the action of a conservative force The work done on the particle during this interval of time is:

Options

A. 0.32 J
B. 6.9 J
C. 9.6 J
D. 0.96 J

Explanation

To solve the problem step by step, we will calculate the work done on the particle as it changes its velocity due to the action of a conservative force.

### Step 1: Identify the given values
- Mass of the particle, \( m = 0.01 \, \text{kg} \)
- Initial velocity, \( \vec{v_i} = 4 \hat{i} + 16 \hat{k} \, \text{m/s} \)
- Final velocity, \( \vec{v_f} = 8 \hat{i} + 20 \hat{j} \, \text{m/s} \)

### Step 2: Calculate the magnitude of the initial velocity
The magnitude of the initial velocity \( v_i \) can be calculated using the formula:
\[ v_i = \sqrt{(4)^2 + (16)^2} = \sqrt{16 + 256} = \sqrt{272} = 16.49 \, \text{m/s} \]

### Step 3: Calculate the magnitude of the final velocity
The magnitude of the final velocity \( v_f \) can be calculated similarly:
\[ v_f = \sqrt{(8)^2 + (20)^2} = \sqrt{64 + 400} = \sqrt{464} = 21.54 \, \text{m/s} \]

### Step 4: Calculate the initial kinetic energy
The initial kinetic energy \( KE_i \) is given by:
\[ KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 0.01 \times (16.49)^2 \]
Calculating \( (16.49)^2 \):
\[ (16.49)^2 \approx 272.56 \]
Thus,
\[ KE_i = \frac{1}{2} \times 0.01 \times 272.56 \approx 1.3628 \, \text{J} \]

### Step 5: Calculate the final kinetic energy
The final kinetic energy \( KE_f \) is given by:
\[ KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 0.01 \times (21.54)^2 \]
Calculating \( (21.54)^2 \):
\[ (21.54)^2 \approx 464.53 \]
Thus,
\[ KE_f = \frac{1}{2} \times 0.01 \times 464.53 \approx 2.32265 \, \text{J} \]

### Step 6: Calculate the work done
The work done \( W \) on the particle is equal to the change in kinetic energy:
\[ W = KE_f - KE_i = 2.32265 - 1.3628 \approx 0.96 \, \text{J} \]

### Final Answer
The work done on the particle during this interval of time is approximately \( 0.96 \, \text{J} \).

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