An object of mass 8 kg starts to move from rest under the action of a variable force `F=3xN` where x is the distance (in m) covered by the object. If initially, the position of the object is x = 2 m, then find its speed `("in m s"^(-1))` when it crosses x = 10 m.

Answer

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.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Mass of the object, \( m = 8 \, \text{kg} \)
- Initial position, \( x_i = 2 \, \text{m} \)
- Final position, \( x_f = 10 \, \text{m} \)
- Variable force, \( F = 3x \, \text{N} \)

2. **Calculate the Work Done**:
The work done by the force as the object moves from \( x = 2 \, \text{m} \) to \( x = 10 \, \text{m} \) can be calculated using the integral of the force over the distance:
\[ W = \int_{x_i}^{x_f} F \, dx = \int_{2}^{10} 3x \, dx \]

3. **Evaluate the Integral**:
Calculate the integral:
\[ W = 3 \int_{2}^{10} x \, dx = 3 \left[ \frac{x^2}{2} \right]_{2}^{10} \]
\[ = 3 \left( \frac{10^2}{2} - \frac{2^2}{2} \right) = 3 \left( \frac{100}{2} - \frac{4}{2} \right) = 3 \left( 50 - 2 \right) = 3 \times 48 = 144 \, \text{J} \]

4. **Apply the Work-Energy Theorem**:
The work done on the object is equal to the change in kinetic energy:
\[ W = \Delta KE = KE_f - KE_i \]
Since the object starts from rest, \( KE_i = 0 \):
\[ W = KE_f = \frac{1}{2} mv^2 \]
Therefore,
\[ 144 = \frac{1}{2} \times 8 \times v^2 \]

5. **Solve for Speed \( v \)**:
Rearranging the equation gives:
\[ 144 = 4v^2 \]
\[ v^2 = \frac{144}{4} = 36 \]
\[ v = \sqrt{36} = 6 \, \text{m/s} \]

### Final Answer:
The speed of the object when it crosses \( x = 10 \, \text{m} \) is \( 6 \, \text{m/s} \).