In the arrangement shown in figure, the track is frictionless until the body reaches the higher level. A frictional force stops the body in a distance d from A. The initial speed of the body `v_(0)` is 6 m/s the height h is 1 m and the coefficient of kinetic friction is 0.2. What is the maximum distance d travelled by body on rough surface?

Options

A. 3 m
B. 4 m
C. 5 m
D. 4.25 m

Answer

To solve the problem step by step, we will use the principles of conservation of energy and the work-energy theorem.

### Step 1: Calculate the final velocity (V) at point A using conservation of energy.
The initial kinetic energy (KE_initial) and potential energy (PE_initial) at the starting point (point B) can be expressed as:
- KE_initial = \( \frac{1}{2} m v_0^2 \)
- PE_initial = 0 (since we are taking the reference level at point B)

At point A, the body has kinetic energy (KE_final) and potential energy (PE_final):
- KE_final = \( \frac{1}{2} m V^2 \)
- PE_final = \( mgh \) (where h is the height)

Using conservation of energy:
\[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \]

Substituting the values:
\[ \frac{1}{2} m v_0^2 + 0 = \frac{1}{2} m V^2 + mgh \]

We can cancel mass (m) from both sides:
\[ \frac{1}{2} v_0^2 = \frac{1}{2} V^2 + gh \]

Now substituting \( v_0 = 6 \, m/s \), \( g = 10 \, m/s^2 \), and \( h = 1 \, m \):
\[ \frac{1}{2} (6^2) = \frac{1}{2} V^2 + 10 \cdot 1 \]
\[ 18 = \frac{1}{2} V^2 + 10 \]

Rearranging gives:
\[ \frac{1}{2} V^2 = 18 - 10 \]
\[ \frac{1}{2} V^2 = 8 \]
\[ V^2 = 16 \]
\[ V = 4 \, m/s \]

### Step 2: Calculate the distance (d) traveled on the rough surface.
The work done against friction (W_friction) is equal to the change in kinetic energy as the body comes to rest:
\[ W_{friction} = KE_{initial} - KE_{final} \]
Since the body comes to rest, \( KE_{final} = 0 \):
\[ W_{friction} = KE_{initial} \]

The initial kinetic energy at point A is:
\[ KE_{initial} = \frac{1}{2} m V^2 \]
Substituting \( V = 4 \, m/s \):
\[ KE_{initial} = \frac{1}{2} m (4^2) = \frac{1}{2} m (16) = 8m \]

The work done against friction can also be expressed as:
\[ W_{friction} = F_{friction} \cdot d \]
Where \( F_{friction} = \mu mg \) (with \( \mu = 0.2 \)):
\[ F_{friction} = 0.2mg \]

Thus, we have:
\[ 8m = 0.2mg \cdot d \]

Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 8 = 0.2g \cdot d \]

Substituting \( g = 10 \, m/s^2 \):
\[ 8 = 0.2 \cdot 10 \cdot d \]
\[ 8 = 2d \]
\[ d = \frac{8}{2} = 4 \, m \]

### Final Answer:
The maximum distance \( d \) traveled by the body on the rough surface is **4 meters**.

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