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, we need to find the value of \( k \) given the conditions of the motion of a body experiencing a frictional force proportional to its velocity. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Forces Acting on the Body
The body of mass \( m = 1 \, \text{kg} \) experiences a frictional force given by:
\[ F = -kv \]
where \( v \) is the velocity of the body and \( k \) is a constant we need to find.

### Step 2: Apply Newton's Second Law
According to Newton's second law:
\[ F = ma \]
where \( a \) is the acceleration. Since \( a = \frac{dv}{dt} \), we can write:
\[ m \frac{dv}{dt} = -kv \]
Substituting \( m = 1 \, \text{kg} \):
\[ \frac{dv}{dt} = -kv \]

### Step 3: Separate Variables and Integrate
We can separate the variables:
\[ \frac{dv}{v} = -k \, dt \]
Now, we integrate both sides. The left side integrates to \( \ln |v| \) and the right side integrates to \( -kt \):
\[ \int \frac{dv}{v} = \int -k \, dt \implies \ln |v| = -kt + C \]
where \( C \) is the constant of integration.

### Step 4: Solve for the Constant of Integration
To find \( C \), we use the initial condition. At \( t = 0 \), \( v = v_0 = 10 \, \text{m/s} \):
\[ \ln |10| = C \implies C = \ln 10 \]
Thus, we have:
\[ \ln |v| = -kt + \ln 10 \]

### Step 5: Exponentiate to Solve for \( v \)
Exponentiating both sides gives:
\[ v = 10 e^{-kt} \]

### Step 6: Find the Kinetic Energy
The initial kinetic energy \( KE_0 \) is given by:
\[ KE_0 = \frac{1}{2} m v_0^2 = \frac{1}{2} \cdot 1 \cdot (10)^2 = 50 \, \text{J} \]
After \( t = 10 \, \text{s} \), the kinetic energy becomes half:
\[ KE = \frac{1}{2} KE_0 = 25 \, \text{J} \]

### Step 7: Set Up the Equation for Kinetic Energy at \( t = 10 \, \text{s} \)
At \( t = 10 \, \text{s} \), the velocity is:
\[ v(10) = 10 e^{-10k} \]
The kinetic energy at this time is:
\[ KE = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 1 \cdot (10 e^{-10k})^2 = 50 e^{-20k} \]
Setting this equal to 25 J:
\[ 50 e^{-20k} = 25 \]

### Step 8: Solve for \( k \)
Dividing both sides by 50:
\[ e^{-20k} = \frac{25}{50} = \frac{1}{2} \]
Taking the natural logarithm:
\[ -20k = \ln\left(\frac{1}{2}\right) \]
Thus:
\[ k = -\frac{1}{20} \ln\left(\frac{1}{2}\right) = \frac{1}{20} \ln(2) \]

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