If the force acting on a body is inversely proportional to its speed, then its kinetic energy is

Options

A. linearly related to time
B. inversely proportional to time
C. inversely proportional to the square of time
D. a constant

Explanation

To solve the problem where the force acting on a body is inversely proportional to its speed, we will derive the relationship between kinetic energy and time step by step.

### Step 1: Understand the relationship between force and speed
Given that the force \( F \) is inversely proportional to speed \( v \), we can express this relationship mathematically as:
\[ F = \frac{k}{v} \]
where \( k \) is a constant of proportionality.

### Step 2: Relate force to acceleration
From Newton's second law, we know that:
\[ F = ma \]
where \( m \) is the mass of the body and \( a \) is its acceleration. Since acceleration \( a \) can be expressed as the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
we can substitute this into the equation for force:
\[ ma = m \frac{dv}{dt} \]

### Step 3: Set up the equation
Now, we can set the two expressions for force equal to each other:
\[ m \frac{dv}{dt} = \frac{k}{v} \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ m v \frac{dv}{dt} = k \]
This can be rewritten as:
\[ m v \, dv = k \, dt \]

### Step 5: Integrate both sides
Now we will integrate both sides. The left side will be integrated with respect to \( v \) from \( 0 \) to \( v \), and the right side will be integrated with respect to \( t \) from \( 0 \) to \( t \):
\[ \int_0^v m v \, dv = \int_0^t k \, dt \]

### Step 6: Perform the integration
The left side integrates to:
\[ m \left[ \frac{v^2}{2} \right]_0^v = \frac{m v^2}{2} \]
The right side integrates to:
\[ k [t]_0^t = kt \]

### Step 7: Equate the results
Setting the results of the integrals equal to each other gives us:
\[ \frac{m v^2}{2} = kt \]

### Step 8: Relate to kinetic energy
The left side of the equation, \( \frac{m v^2}{2} \), is the expression for kinetic energy \( KE \):
\[ KE = \frac{m v^2}{2} \]
Thus, we have:
\[ KE = kt \]

### Conclusion
This shows that the kinetic energy \( KE \) is linearly related to time \( t \) since \( k \) is a constant.

### Final Answer
If the force acting on a body is inversely proportional to its speed, then its kinetic energy is linearly related to time.

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