Consider a frictionless ramp on which a smooth object is made to slide down from an initial height 'h'. The distance 'd' necessary to stop the object on a flat track of coefficient of friction '`mu`.'), kept at the ramp end is

Options

A. `h//mu`
B. `mu h`
C. `mu^(2) h`
D. `h^(2) mu`

Answer

To solve the problem, we need to determine the distance 'd' that an object will slide on a flat track with a coefficient of friction 'μ' after sliding down a frictionless ramp from an initial height 'h'.

### Step-by-Step Solution:

1. **Identify the Initial Energy**:
When the object is at height 'h', it has gravitational potential energy given by:
\[ PE = mgh \]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.

2. **Energy Conversion**:
As the object slides down the ramp, this potential energy is converted into kinetic energy (KE) at the bottom of the ramp. The kinetic energy at the bottom is given by:
\[ KE = \frac{1}{2} mv^2 \]
At the bottom of the ramp, all potential energy has been converted into kinetic energy:
\[ mgh = \frac{1}{2} mv^2 \]

3. **Solve for Velocity**:
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{1}{2} v^2 \implies v^2 = 2gh \implies v = \sqrt{2gh} \]

4. **Determine the Work Done by Friction**:
When the object reaches the flat surface, it will experience a frictional force that does work against its motion. The frictional force \( F_f \) is given by:
\[ F_f = \mu mg \]
The work done by friction \( W_f \) over the distance 'd' is:
\[ W_f = -F_f \cdot d = -\mu mgd \]
(The negative sign indicates that the work done by friction is in the opposite direction of the motion.)

5. **Apply the Work-Energy Principle**:
According to the work-energy theorem, the work done by friction will equal the change in kinetic energy:
\[ W_f = \Delta KE \]
Since the object comes to a stop, the final kinetic energy is zero, and the initial kinetic energy is \( \frac{1}{2} mv^2 \):
\[ -\mu mgd = 0 - \frac{1}{2} mv^2 \]
Substituting \( v^2 = 2gh \):
\[ -\mu mgd = -\frac{1}{2} m(2gh) \]

6. **Simplify the Equation**:
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu gd = gh \]
Rearranging gives:
\[ d = \frac{gh}{\mu g} = \frac{h}{\mu} \]

### Final Answer:
Thus, the distance 'd' necessary to stop the object on the flat track is:
\[ d = \frac{h}{\mu} \]