Consider the body of mass 0.2 kg and a smooth incline of hight 3.2 m and length 10 m. Select the correct alternative

Options

A. a. Minimum work required to lift the block from the ground and put it at the top is 6.4 J
B. b. Work required to slide the body up the incline (slowly) is 6.4 J
C. c. Maximum speed of the body slipping down from rest on the plane, on reaching the ground is 8 m/s
D. d. Work required to slide the body down the plane is more than 6.4 J

Answer

To solve the problem step by step, we will analyze the given information and calculate the required values.

### Given:
- Mass of the body, \( m = 0.2 \, \text{kg} \)
- Height of the incline, \( h = 3.2 \, \text{m} \)
- Length of the incline, \( L = 10 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 1: Calculate the work done to lift the block to the top of the incline.
The work done against gravity to lift the block is equal to the change in potential energy.

\[ \text{Work done} = \Delta PE = mgh \]

Substituting the values:

\[ \text{Work done} = 0.2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 3.2 \, \text{m} \]

Calculating this:

\[ \text{Work done} = 0.2 \times 10 \times 3.2 = 6.4 \, \text{J} \]

### Step 2: Work required to slide the body up the incline slowly.
When sliding up slowly, the work done against gravity is the same as lifting it directly:

\[ \text{Work required} = 6.4 \, \text{J} \]

### Step 3: Calculate the maximum speed of the body slipping down the incline.
When the body slips down the incline, all the potential energy converts into kinetic energy at the bottom of the incline.

Using the energy conservation principle:

\[ \Delta PE = KE \]

Where \( KE = \frac{1}{2} mv^2 \). Thus,

\[ mgh = \frac{1}{2} mv^2 \]

Cancelling \( m \) from both sides (since \( m \neq 0 \)):

\[ gh = \frac{1}{2} v^2 \]

Rearranging gives:

\[ v^2 = 2gh \]

Substituting the values:

\[ v^2 = 2 \times 10 \, \text{m/s}^2 \times 3.2 \, \text{m} = 64 \]

Taking the square root:

\[ v = \sqrt{64} = 8 \, \text{m/s} \]

### Step 4: Work required to slide the body down the plane.
Since the incline is smooth (frictionless), the work done to slide down is equal to the potential energy lost, which is:

\[ \text{Work down} = mgh = 6.4 \, \text{J} \]

### Conclusion:
- The minimum work required to lift the block to the top is \( 6.4 \, \text{J} \) (Correct).
- The work required to slide the body up slowly is also \( 6.4 \, \text{J} \) (Correct).
- The maximum speed of the body slipping down is \( 8 \, \text{m/s} \) (Correct).
- The work required to slide the body down the plane is not more than \( 6.4 \, \text{J} \) (Incorrect).

### Final Answer:
Options A, B, and C are correct. Option D is incorrect.

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