The velocity of a block of mass `2kg` moving along `x-` axis at any time `t` is givne by `v=20-10t(m//s)` where `t` is in seconds and `v` is in `m//s`. At time `t=0`, the block is moving in positive `x-` direction.
The Kinetic energy of block at `t=3 sec.` is `:`

Options

A. `50J`
B. `100J`
C. `200J`
D. `300J`

Explanation

To solve the problem, we need to calculate the kinetic energy of a block of mass 2 kg at time \( t = 3 \) seconds, given its velocity as a function of time.

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of the block, \( m = 2 \, \text{kg} \)
- Velocity as a function of time, \( v(t) = 20 - 10t \, \text{m/s} \)
- Time at which we need to find the kinetic energy, \( t = 3 \, \text{s} \)

2. **Calculate the velocity at \( t = 3 \) seconds:**
\[ v(3) = 20 - 10 \times 3 \]
\[ v(3) = 20 - 30 = -10 \, \text{m/s} \]

3. **Use the kinetic energy formula:**
The kinetic energy \( K \) is given by the formula:
\[ K = \frac{1}{2} m v^2 \]

4. **Substitute the values into the kinetic energy formula:**
\[ K = \frac{1}{2} \times 2 \times (-10)^2 \]
\[ K = 1 \times 100 = 100 \, \text{J} \]

5. **Conclusion:**
The kinetic energy of the block at \( t = 3 \) seconds is \( 100 \, \text{J} \).