A body of mass `m=10^(-2) kg` is moving in a medium and experiences a frictional force `F=-Kupsilon^(2).` Its initial speed is `upsilon_(0)=10ms^(-2)`. If , after `10s`, its energy is `(1)/(8)m upsilon_(0)^(2)`, the value of `k` will be

Options

A. 1/10
B. 1/10000
C. 1000
D. 100

Answer

To solve the problem, we need to find the value of \( k \) given the mass \( m = 10^{-2} \, \text{kg} \), the initial speed \( v_0 = 10 \, \text{m/s} \), and the final energy after \( t = 10 \, \text{s} \) is \( \frac{1}{8} m v_0^2 \).

### Step-by-Step Solution:

1. **Calculate Initial Kinetic Energy**:
The initial kinetic energy \( E_k \) of the body can be calculated using the formula:
\[ E_k = \frac{1}{2} m v_0^2 \]
Substituting the values:
\[ E_k = \frac{1}{2} \times 10^{-2} \, \text{kg} \times (10 \, \text{m/s})^2 = \frac{1}{2} \times 10^{-2} \times 100 = 0.5 \times 10^{-2} = 5 \times 10^{-3} \, \text{J} \]

2. **Final Energy After 10 Seconds**:
According to the problem, the energy after \( 10 \, \text{s} \) is given by:
\[ E_f = \frac{1}{8} m v_0^2 \]
Substituting the values:
\[ E_f = \frac{1}{8} \times 10^{-2} \times (10)^2 = \frac{1}{8} \times 10^{-2} \times 100 = \frac{1}{8} \times 10^{-2} \times 100 = \frac{1}{8} \times 10^{-2} \times 100 = 1.25 \times 10^{-2} \, \text{J} \]

3. **Determine Change in Kinetic Energy**:
The change in kinetic energy \( \Delta E \) over \( 10 \, \text{s} \) is:
\[ \Delta E = E_k - E_f = 5 \times 10^{-3} - 1.25 \times 10^{-2} = -7.5 \times 10^{-3} \, \text{J} \]

4. **Relate Frictional Force to Energy Loss**:
The work done by the frictional force over time can be expressed as:
\[ W = F \cdot d \]
Since the frictional force is given as \( F = -k v^2 \), we can relate it to the energy lost. The energy lost due to friction is equal to the change in kinetic energy:
\[ -\Delta E = k \int_0^{10} v^2 \, dt \]

5. **Using the Average Velocity**:
Since the body slows down, we can use the average velocity to estimate the distance traveled. The average velocity \( v_{avg} \) can be calculated as:
\[ v_{avg} = \frac{v_0 + v_f}{2} \]
where \( v_f \) is the final velocity. From the energy relation, we found \( v_f = \frac{v_0}{2} = 5 \, \text{m/s} \).
\[ v_{avg} = \frac{10 + 5}{2} = 7.5 \, \text{m/s} \]
The distance \( d \) traveled in \( 10 \, \text{s} \) is:
\[ d = v_{avg} \cdot t = 7.5 \, \text{m/s} \cdot 10 \, \text{s} = 75 \, \text{m} \]

6. **Calculate Work Done by Friction**:
The work done by the frictional force is:
\[ W = -\Delta E = 7.5 \times 10^{-3} \, \text{J} \]
Therefore,
\[ W = k \cdot d \implies 7.5 \times 10^{-3} = k \cdot 75 \]
Solving for \( k \):
\[ k = \frac{7.5 \times 10^{-3}}{75} = 10^{-4} \, \text{kg/m} \]

### Final Answer:
The value of \( k \) is \( 10^{-4} \, \text{kg/m} \).