A force F=-4x-8 (in N) is acting on a block where x is the position of the block in metres. The energy of oscillation is 32 J. The block oscillates between two points, out of which the value of position of one point (in metres) is an integer from 0 to 9. Find it.

Explanation

To solve the problem step by step, we will analyze the given force, derive the necessary parameters, and find the required position of the block.

### Step 1: Identify the Force Equation
The force acting on the block is given as:
\[ F = -4x - 8 \]

### Step 2: Rewrite the Force Equation
We can factor out the common term:
\[ F = -4(x + 2) \]
This indicates that the force is a linear function of position \( x \) and has a mean position (equilibrium position) at \( x = -2 \).

### Step 3: Relate Force to Simple Harmonic Motion (SHM)
In SHM, the force can be expressed as:
\[ F = -kx \]
where \( k \) is the spring constant. From our equation, we can see that:
\[ k = 4 \]

### Step 4: Calculate the Amplitude from Energy
The total mechanical energy \( E \) in SHM is given by:
\[ E = \frac{1}{2} k A^2 \]
where \( A \) is the amplitude. We know the energy is given as 32 J:
\[ 32 = \frac{1}{2} \times 4 \times A^2 \]

### Step 5: Solve for Amplitude
Rearranging the equation:
\[ 32 = 2A^2 \]
\[ A^2 = \frac{32}{2} = 16 \]
\[ A = 4 \]

### Step 6: Determine the Extreme Positions
The mean position is at \( x = -2 \), and the amplitude is \( 4 \). Therefore, the extreme positions (maximum and minimum) can be calculated as:
- Maximum position: \( -2 + 4 = 2 \)
- Minimum position: \( -2 - 4 = -6 \)

### Step 7: Identify the Range of Oscillation
The block oscillates between \( -6 \) and \( 2 \). We need to find the integer position within the range of \( 0 \) to \( 9 \).

### Step 8: Conclusion
The integer position from \( 0 \) to \( 9 \) that the block oscillates to is:
\[ \text{Position} = 2 \]

Thus, the final answer is:
**The integer position is 2.**

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