A block of mass 2 kg is having velocity `4sqrt(5) ms^(-1)` in the positive x - direction at the origin. The only force acting on it is `F = (3x^2 - 12 x)N`. Its velocity when it is at `x = 2` m is

Options

A. `8 m s^(-1)`
B. `4 ms^(-1)`
C. `10 sqrt(24) ms^(-1)`
D. `20 ms^(-1)`

Answer

To find the velocity of the block when it is at \( x = 2 \) m, we will use the work-energy principle, which states that the work done by the forces acting on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the block, \( m = 2 \, \text{kg} \)
- Initial velocity at \( x = 0 \), \( v_0 = 4\sqrt{5} \, \text{m/s} \)
- Force acting on the block, \( F = 3x^2 - 12x \, \text{N} \)

2. **Calculate the Initial Kinetic Energy:**
\[ KE_0 = \frac{1}{2} m v_0^2 = \frac{1}{2} \times 2 \times (4\sqrt{5})^2 = 1 \times 80 = 80 \, \text{J} \]

3. **Calculate the Work Done by the Force from \( x = 0 \) to \( x = 2 \):**
The work done \( W \) can be calculated by integrating the force over the displacement:
\[ W = \int_{0}^{2} (3x^2 - 12x) \, dx \]

- Calculate the integral:
\[ W = \left[ x^3 - 6x^2 \right]_{0}^{2} = \left[ 2^3 - 6(2^2) \right] - \left[ 0 - 0 \right] = 8 - 24 = -16 \, \text{J} \]

4. **Calculate the Final Kinetic Energy:**
The work-energy principle states:
\[ KE_f = KE_0 + W \]
\[ KE_f = 80 \, \text{J} - 16 \, \text{J} = 64 \, \text{J} \]

5. **Calculate the Final Velocity:**
Using the final kinetic energy to find the final velocity \( v_f \):
\[ KE_f = \frac{1}{2} m v_f^2 \]
\[ 64 = \frac{1}{2} \times 2 \times v_f^2 \implies 64 = v_f^2 \implies v_f = \sqrt{64} = 8 \, \text{m/s} \]

### Final Answer:
The velocity of the block when it is at \( x = 2 \, \text{m} \) is \( 8 \, \text{m/s} \).