A block of mass 100g is moved with a speed of 5.0 m/s at the highest point in a closed circular tube of radius 10 cm kept in a vertical plane. The cross sectiion of th tube is such that the block just fits in it. The block makes wseverl oscillations inside the tube and finally stops at the lowest point. find teh work done by the tube on the block during the process

Explanation

To solve the problem, we need to find the work done by the tube on the block as it moves from the highest point to the lowest point inside the tube. Here's a step-by-step breakdown of the solution:

### Step 1: Convert the mass of the block to kilograms
The mass of the block is given as 100 grams. To convert this to kilograms:
\[ \text{Mass} = 100 \, \text{g} = 0.1 \, \text{kg} \]

**Hint:** Remember that 1 kg = 1000 g.

### Step 2: Calculate the initial kinetic energy
The initial speed of the block is given as 5.0 m/s. The formula for kinetic energy (KE) is:
\[ \text{KE} = \frac{1}{2} mv^2 \]
Substituting the values:
\[ \text{KE}_{\text{initial}} = \frac{1}{2} \times 0.1 \, \text{kg} \times (5.0 \, \text{m/s})^2 = \frac{1}{2} \times 0.1 \times 25 = 1.25 \, \text{J} \]

**Hint:** Ensure you square the velocity before multiplying by mass.

### Step 3: Determine the final kinetic energy
At the lowest point, the block comes to rest, so the final kinetic energy is:
\[ \text{KE}_{\text{final}} = 0 \, \text{J} \]

**Hint:** The block stops, which means its speed is 0 m/s.

### Step 4: Calculate the change in kinetic energy
The change in kinetic energy (ΔKE) is given by:
\[ \Delta KE = \text{KE}_{\text{final}} - \text{KE}_{\text{initial}} = 0 - 1.25 = -1.25 \, \text{J} \]

**Hint:** The change in energy can be negative if the final energy is less than the initial energy.

### Step 5: Calculate the work done by gravity
The work done by gravity (W_gravity) when moving down a distance equal to twice the radius (2r) is given by:
\[ W_{\text{gravity}} = mgh \]
Where \( h = 2r \) and \( r = 0.1 \, \text{m} \):
\[ h = 2 \times 0.1 = 0.2 \, \text{m} \]
Thus,
\[ W_{\text{gravity}} = 0.1 \, \text{kg} \times 10 \, \text{m/s}^2 \times 0.2 \, \text{m} = 0.1 \times 10 \times 0.2 = 0.2 \, \text{J} \]

**Hint:** Use \( g = 10 \, \text{m/s}^2 \) for simplicity in calculations.

### Step 6: Apply the work-energy principle
According to the work-energy principle:
\[ W_{\text{tube}} + W_{\text{gravity}} = \Delta KE \]
Rearranging gives:
\[ W_{\text{tube}} = \Delta KE - W_{\text{gravity}} \]
Substituting the values:
\[ W_{\text{tube}} = -1.25 \, \text{J} - 0.2 \, \text{J} = -1.45 \, \text{J} \]

**Hint:** The work done by the tube is negative, indicating that it is doing work against the motion of the block.

### Final Answer
The work done by the tube on the block during the process is:
\[ \boxed{-1.45 \, \text{J}} \]