A block of mass 1 slides down a curved track that is one quadrant of a circle of radius 1m. Its speed at the bottom is 2m/s. The work done by frictional force is:

Options

A. 8j
B. `-8J`
C. `4J`
D. `-4J`

Answer

To solve the problem, we will use the principle of conservation of energy. The work done by friction can be calculated by comparing the initial potential energy of the block at the top of the track with its final kinetic energy at the bottom.

### Step-by-Step Solution:

1. **Identify the initial potential energy (PE)**:
The block is at a height (h) equal to the radius of the circular track, which is 1 meter. The potential energy at the top can be calculated using the formula:
\[ PE = mgh \]
where:
- \( m = 1 \, \text{kg} \) (mass of the block)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 1 \, \text{m} \) (height)

Substituting the values:
\[ PE = 1 \times 10 \times 1 = 10 \, \text{J} \]

2. **Calculate the final kinetic energy (KE)**:
The kinetic energy at the bottom of the track can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \( v = 2 \, \text{m/s} \) (speed at the bottom)

Substituting the values:
\[ KE = \frac{1}{2} \times 1 \times (2)^2 = \frac{1}{2} \times 1 \times 4 = 2 \, \text{J} \]

3. **Apply the work-energy principle**:
The work done by friction (W_friction) can be found by comparing the initial potential energy and the final kinetic energy:
\[ W_{\text{friction}} = KE - PE \]
Substituting the values we found:
\[ W_{\text{friction}} = 2 \, \text{J} - 10 \, \text{J} = -8 \, \text{J} \]

4. **Conclusion**:
The work done by frictional forces is:
\[ W_{\text{friction}} = -8 \, \text{J} \]

### Final Answer:
The work done by frictional force is \(-8 \, \text{J}\).

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