If a body has kinetic energy. T, moving on a rough horizontal surface stops at distance y. The frictional force exerted on the body is

Options

A. `T/sqrty`
B. `sqrtT/y`
C. `yT`
D. `T/y`

Answer

To solve the problem, we need to determine the frictional force exerted on a body that has kinetic energy \( T \) and comes to a stop after traveling a distance \( y \) on a rough horizontal surface.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
The kinetic energy \( T \) of a body in motion is given by the formula:
\[ T = \frac{1}{2} m u^2 \]
where \( m \) is the mass of the body and \( u \) is its initial velocity.

2. **Using the Equation of Motion**:
When the body comes to a stop after traveling a distance \( y \), we can use the equation of motion:
\[ v^2 = u^2 + 2as \]
Here, \( v \) is the final velocity (which is 0 when the body stops), \( u \) is the initial velocity, \( a \) is the acceleration (retardation in this case), and \( s \) is the distance traveled (which is \( y \)).

3. **Setting Up the Equation**:
Since the final velocity \( v = 0 \), we can rearrange the equation:
\[ 0 = u^2 + 2(-a)y \]
This simplifies to:
\[ u^2 = 2ay \]

4. **Relating Kinetic Energy and Frictional Force**:
From the kinetic energy equation, we can express \( u^2 \) in terms of \( T \):
\[ u^2 = \frac{2T}{m} \]
Now, we can substitute this expression for \( u^2 \) into our previous equation:
\[ \frac{2T}{m} = 2ay \]

5. **Solving for Acceleration \( a \)**:
Dividing both sides by 2:
\[ \frac{T}{m} = ay \]
Rearranging gives us:
\[ a = \frac{T}{my} \]

6. **Finding the Frictional Force**:
The frictional force \( F \) can be expressed using Newton's second law:
\[ F = ma \]
Substituting the expression for \( a \):
\[ F = m \left( \frac{T}{my} \right) \]
Simplifying this, we find:
\[ F = \frac{T}{y} \]

### Final Answer:
The frictional force exerted on the body is:
\[ F = \frac{T}{y} \]