A particle of mass 5 kg moves in a circle of radius 20 cm. Its linear speed at a time t is given by v=4t, t is in s and v is in `ms^(-1)` The net force acting on the particle at 1 = 0.5 s

Options

A. `20 sqrt(28)N`
B. `20 N`
C. `100 N`
D. `20 sqrt(26) N`

Explanation

To solve the problem, we need to find the net force acting on a particle of mass 5 kg moving in a circular path with a radius of 20 cm at a time \( t = 0.5 \) s. The linear speed of the particle is given by \( v = 4t \).

### Step-by-Step Solution:

1. **Convert the radius to meters**:
\[ r = 20 \text{ cm} = 0.2 \text{ m} \]

2. **Calculate the linear speed at \( t = 0.5 \) s**:
\[ v = 4t = 4 \times 0.5 = 2 \text{ m/s} \]

3. **Calculate the tangential acceleration**:
The tangential acceleration \( a_t \) can be found by differentiating the velocity with respect to time:
\[ a_t = \frac{dv}{dt} = 4 \text{ m/s}^2 \]

4. **Calculate the centripetal acceleration**:
The centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = \frac{v^2}{r} \]
Substituting the values:
\[ a_c = \frac{(2 \text{ m/s})^2}{0.2 \text{ m}} = \frac{4}{0.2} = 20 \text{ m/s}^2 \]

5. **Calculate the net acceleration**:
The net acceleration \( a_{net} \) is the vector sum of the tangential and centripetal accelerations:
\[ a_{net} = \sqrt{a_c^2 + a_t^2} \]
Substituting the values:
\[ a_{net} = \sqrt{(20 \text{ m/s}^2)^2 + (4 \text{ m/s}^2)^2} = \sqrt{400 + 16} = \sqrt{416} = 4\sqrt{26} \text{ m/s}^2 \]

6. **Calculate the net force**:
Using Newton's second law, the net force \( F_{net} \) is given by:
\[ F_{net} = m \cdot a_{net} \]
Substituting the mass and net acceleration:
\[ F_{net} = 5 \text{ kg} \cdot 4\sqrt{26} \text{ m/s}^2 = 20\sqrt{26} \text{ N} \]

### Final Answer:
The net force acting on the particle at \( t = 0.5 \) s is:
\[ F_{net} = 20\sqrt{26} \text{ N} \]