A block of mass m is gently placed over a massive plank moving horizontal over a smooth surface with velocity `10 ms^(-1)`. The coefficient of friction between the block and the plank is 0.2. The distance travelled by the block till it slides on the plank is `[g = 10 ms^(-2)]`

Options

A. 10 m
B. 15m
C. 25 m
D. 35m

Answer

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

### Step 1: Identify the given values
- Mass of the block, \( m = 2 \, \text{kg} \) (assumed from the transcript)
- Coefficient of friction, \( \mu = 0.2 \)
- Initial velocity of the plank (and block), \( v = 10 \, \text{ms}^{-1} \)
- Acceleration due to gravity, \( g = 10 \, \text{ms}^{-2} \)

### Step 2: Calculate the initial kinetic energy of the block
The initial kinetic energy (\( KE_i \)) of the block is given by the formula:
\[ KE_i = \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE_i = \frac{1}{2} \times 2 \, \text{kg} \times (10 \, \text{ms}^{-1})^2 = \frac{1}{2} \times 2 \times 100 = 100 \, \text{J} \]

### Step 3: Calculate the frictional force acting on the block
The frictional force (\( F_r \)) can be calculated using the formula:
\[ F_r = \mu mg \]
Substituting the values:
\[ F_r = 0.2 \times 2 \, \text{kg} \times 10 \, \text{ms}^{-2} = 0.2 \times 20 = 4 \, \text{N} \]

### Step 4: Apply the work-energy theorem
According to the work-energy theorem, the work done by the frictional force is equal to the change in kinetic energy:
\[ \text{Work done by friction} = KE_f - KE_i \]
Since the block comes to rest, \( KE_f = 0 \):
\[ -KE_i = -F_r \cdot d \]
Where \( d \) is the distance traveled by the block until it slides. Rearranging gives:
\[ F_r \cdot d = KE_i \]
Substituting the values we have:
\[ 4 \, \text{N} \cdot d = 100 \, \text{J} \]

### Step 5: Solve for distance \( d \)
\[ d = \frac{100 \, \text{J}}{4 \, \text{N}} = 25 \, \text{m} \]

### Conclusion
The distance traveled by the block until it slides on the plank is \( 25 \, \text{m} \).

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