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 step-by-step, we will follow these steps:

### Step 1: Identify the given information
- Mass of the object, \( m = 8 \, \text{kg} \)
- Force acting on the object, \( F = 3x \, \text{N} \)
- Initial position, \( x_i = 2 \, \text{m} \)
- Final position, \( x_f = 10 \, \text{m} \)
- Initial velocity, \( v_i = 0 \, \text{m/s} \)

### Step 2: Relate force to acceleration
From Newton's second law, we know that:
\[ F = ma \]
Thus, we can express acceleration \( a \) as:
\[ a = \frac{F}{m} = \frac{3x}{8} \]

### Step 3: Use the relationship between acceleration, velocity, and position
Using the relationship:
\[ a = v \frac{dv}{dx} \]
we can substitute for \( a \):
\[ \frac{3x}{8} = v \frac{dv}{dx} \]

### Step 4: Rearranging and integrating
Rearranging gives:
\[ v \, dv = \frac{3x}{8} \, dx \]
Now, we will integrate both sides. The left side will be integrated from \( 0 \) to \( v \) and the right side from \( 2 \) to \( 10 \):
\[ \int_0^v v \, dv = \int_2^{10} \frac{3x}{8} \, dx \]

### Step 5: Perform the integration
The left side integrates to:
\[ \frac{v^2}{2} \bigg|_0^v = \frac{v^2}{2} \]
The right side integrates as follows:
\[ \frac{3}{8} \int_2^{10} x \, dx = \frac{3}{8} \left[ \frac{x^2}{2} \right]_2^{10} = \frac{3}{8} \left( \frac{10^2}{2} - \frac{2^2}{2} \right) = \frac{3}{8} \left( \frac{100}{2} - \frac{4}{2} \right) = \frac{3}{8} \left( 50 - 2 \right) = \frac{3}{8} \times 48 = 18 \]

### Step 6: Set the integrals equal to each other
Now we have:
\[ \frac{v^2}{2} = 18 \]
Multiplying both sides by 2 gives:
\[ v^2 = 36 \]

### Step 7: Solve for \( v \)
Taking the square root:
\[ v = 6 \, \text{m/s} \]

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