A body of mass m = 1 kg is moving in a medium and experiences a frictional force F = -kv, where v is the speed of the body. The initial speed is `v_(0) = 10 ms^(-1)` and after 10 s, its energy becomes half of the initial energy. Then, the value of k is

Options

A. `10 lnsqrt(2)`
B. `ln sqrt(2)`
C. `(1)/(20)ln2`
D. 10 ln 2

Answer

To solve the problem step by step, we start with the given information and apply the principles of physics related to motion and energy.

### Step 1: Understand the initial conditions
The mass of the body \( m = 1 \, \text{kg} \) and the initial speed \( v_0 = 10 \, \text{m/s} \). The initial kinetic energy \( KE_i \) can be calculated using the formula:
\[ KE_i = \frac{1}{2} m v_0^2 = \frac{1}{2} \times 1 \times (10)^2 = 50 \, \text{J} \]

### Step 2: Determine the final kinetic energy
According to the problem, after 10 seconds, the kinetic energy becomes half of the initial energy:
\[ KE_f = \frac{1}{2} KE_i = \frac{1}{2} \times 50 = 25 \, \text{J} \]

### Step 3: Relate kinetic energy to final speed
The final kinetic energy \( KE_f \) can also be expressed in terms of the final speed \( v_f \):
\[ KE_f = \frac{1}{2} m v_f^2 \]
Setting this equal to the value we found:
\[ 25 = \frac{1}{2} \times 1 \times v_f^2 \]
Solving for \( v_f^2 \):
\[ v_f^2 = 50 \implies v_f = \sqrt{50} = 5\sqrt{2} \, \text{m/s} \]

### Step 4: Apply Newton's second law
The frictional force acting on the body is given by \( F = -kv \). According to Newton's second law, \( F = m \frac{dv}{dt} \):
\[ -kv = m \frac{dv}{dt} \]
Substituting \( m = 1 \):
\[ -kv = \frac{dv}{dt} \]

### Step 5: Rearranging and integrating
Rearranging gives:
\[ \frac{dv}{v} = -k \, dt \]
Integrating both sides from \( v_0 = 10 \) to \( v_f = 5\sqrt{2} \) and from \( t = 0 \) to \( t = 10 \):
\[ \int_{10}^{5\sqrt{2}} \frac{dv}{v} = -k \int_{0}^{10} dt \]
This results in:
\[ \ln(v) \bigg|_{10}^{5\sqrt{2}} = -10k \]
Calculating the left side:
\[ \ln(5\sqrt{2}) - \ln(10) = \ln\left(\frac{5\sqrt{2}}{10}\right) = \ln\left(\frac{\sqrt{2}}{2}\right) = \ln\left(\frac{1}{\sqrt{2}}\right) = -\frac{1}{2} \ln(2) \]
Thus, we have:
\[ -\frac{1}{2} \ln(2) = -10k \]
Solving for \( k \):
\[ k = \frac{1}{20} \ln(2) \]

### Final Answer
The value of \( k \) is:
\[ k = \frac{1}{20} \ln(2) \, \text{s}^{-1} \]