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 find the frictional force exerted on a body that has an initial 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 the body is given by the formula:
\[ T = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the body and \( v \) is its initial velocity.

2. **Relating Kinetic Energy to Velocity**:
From the kinetic energy formula, we can express the initial velocity \( v \):
\[ v = \sqrt{\frac{2T}{m}} \]

3. **Using the Work-Energy Principle**:
The work done by the frictional force \( F \) over the distance \( y \) is equal to the change in kinetic energy. Since the body comes to rest, the final kinetic energy is 0. Therefore, the work done by friction is:
\[ W = F \cdot y = \Delta KE = 0 - T = -T \]
This implies:
\[ F \cdot y = -T \]

4. **Finding the Frictional Force**:
Rearranging the equation gives us:
\[ F = -\frac{T}{y} \]
Since the frictional force acts in the opposite direction to the motion, we can express the frictional force as:
\[ F = \frac{T}{y} \]

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