A particle of mass 1 kg moves in a circular path of radius 1 m such that its speed varies with time as per equation `v=3t^(2)m//s` where time t is in seconds. The power delivered by the force acting on the paritcle at `t=1s`, is `:-`

Options

A. 36 W
B. 20 W
C. 18 W
D. zero

Answer

To solve the problem, we need to find the power delivered by the force acting on a particle of mass 1 kg moving in a circular path of radius 1 m, where the speed varies with time according to the equation \( v = 3t^2 \) m/s. We will use the work-energy theorem to find the power.

### Step-by-Step Solution:

1. **Identify the given parameters:**
- Mass of the particle, \( m = 1 \, \text{kg} \)
- Speed of the particle as a function of time, \( v(t) = 3t^2 \, \text{m/s} \)

2. **Calculate the kinetic energy (KE) of the particle:**
The kinetic energy of a particle is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the expression for \( v \):
\[ KE = \frac{1}{2} m (3t^2)^2 = \frac{1}{2} m (9t^4) = \frac{9}{2} mt^4 \]
Since \( m = 1 \, \text{kg} \):
\[ KE = \frac{9}{2} t^4 \]

3. **Differentiate the kinetic energy with respect to time to find power:**
Power \( P \) is defined as the rate of change of work done (or kinetic energy in this case):
\[ P = \frac{d(KE)}{dt} \]
Now, differentiating \( KE = \frac{9}{2} t^4 \):
\[ P = \frac{d}{dt} \left( \frac{9}{2} t^4 \right) = \frac{9}{2} \cdot 4t^3 = 18t^3 \]

4. **Calculate the power at \( t = 1 \, \text{s} \):**
Substitute \( t = 1 \) into the power equation:
\[ P(t=1) = 18(1)^3 = 18 \, \text{W} \]

### Final Answer:
The power delivered by the force acting on the particle at \( t = 1 \, \text{s} \) is \( 18 \, \text{W} \).

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