A vehicle of mass m is moving on a rough horizontal road with kinetic energy 'E'. If the co-efficient of friction between the tyres and the road be `mu`. Then the stopping distance is,

Options

A. `(E)/(2 mu mg)`
B. `(E^(2))/(2 mu mg)`
C. `(E)/(2 mu m^2 g)`
D. `(E)/(mu mg)`

Answer

To solve the problem of finding the stopping distance \( S \) for a vehicle of mass \( m \) moving with kinetic energy \( E \) on a rough horizontal road with a coefficient of friction \( \mu \), we can follow these steps:

### Step 1: Relate Kinetic Energy to Velocity
The kinetic energy \( E \) of the vehicle can be expressed in terms of its mass \( m \) and velocity \( v \):
\[ E = \frac{1}{2} mv^2 \]
From this, we can solve for the velocity \( v \):
\[ v^2 = \frac{2E}{m} \]

### Step 2: Determine the Frictional Force
The frictional force \( F_f \) that will act to stop the vehicle is given by:
\[ F_f = \mu N \]
where \( N \) is the normal force. On a horizontal road, the normal force \( N \) is equal to the weight of the vehicle:
\[ N = mg \]
Thus, the frictional force becomes:
\[ F_f = \mu mg \]

### Step 3: Apply Newton's Second Law
According to Newton's second law, the acceleration \( a \) (which will be negative since it is deceleration) can be expressed as:
\[ F = ma \]
Substituting the frictional force into this equation gives:
\[ -\mu mg = ma \]
From this, we can solve for the acceleration:
\[ a = -\mu g \]

### Step 4: Use the Kinematic Equation
We can now use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2aS \]
Here, the final velocity \( v = 0 \) (when the vehicle stops), the initial velocity \( u = v \), and \( a = -\mu g \). Plugging these values into the equation gives:
\[ 0 = \frac{2E}{m} + 2(-\mu g)S \]
Rearranging this equation to solve for \( S \):
\[ 2\mu g S = \frac{2E}{m} \]
\[ S = \frac{E}{\mu mg} \]

### Final Result
Thus, the stopping distance \( S \) is given by:
\[ S = \frac{E}{\mu mg} \]