To solve the problem, we need to find the value of \( K \) given the conditions of the particle's motion. Let's break down the solution step by step.
### Step 1: Understand the Given Information
- Mass of the particle, \( m = 200 \, \text{g} = 0.2 \, \text{kg} \)
- Radius of the circular path, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Tangential acceleration, \( a_t = 6t \, \text{m/s}^2 \)
- Average power delivered in the first 10 seconds, \( P_{\text{avg}} = 100 \, \text{kW} = 100,000 \, \text{W} \)
- The particle starts from rest.
### Step 2: Calculate the Tangential Force
The tangential force \( F_t \) acting on the particle can be calculated using Newton's second law:
\[
F_t = m \cdot a_t = m \cdot (6t) = 0.2 \cdot (6t) = 1.2t \, \text{N}
\]
### Step 3: Determine the Velocity as a Function of Time
Since the tangential acceleration is given as \( a_t = 6t \), we can find the velocity \( v \) as a function of time by integrating the acceleration:
\[
a_t = \frac{dv}{dt} \implies dv = 6t \, dt
\]
Integrating both sides from \( 0 \) to \( v \) and \( 0 \) to \( t \):
\[
\int_0^v dv = \int_0^t 6t \, dt
\]
This gives:
\[
v = 3t^2
\]
### Step 4: Calculate the Instantaneous Power
The instantaneous power \( P \) delivered to the particle is given by:
\[
P = F_t \cdot v
\]
Substituting \( F_t \) and \( v \):
\[
P = (1.2t) \cdot (3t^2) = 3.6t^3 \, \text{W}
\]
### Step 5: Calculate the Average Power over the First 10 Seconds
The average power over the time interval from \( 0 \) to \( 10 \) seconds is given by:
\[
P_{\text{avg}} = \frac{1}{T} \int_0^T P \, dt
\]
where \( T = 10 \, \text{s} \):
\[
P_{\text{avg}} = \frac{1}{10} \int_0^{10} 3.6t^3 \, dt
\]
Calculating the integral:
\[
\int_0^{10} 3.6t^3 \, dt = 3.6 \cdot \left[\frac{t^4}{4}\right]_0^{10} = 3.6 \cdot \frac{10^4}{4} = 3.6 \cdot 2500 = 9000 \, \text{W}
\]
Thus,
\[
P_{\text{avg}} = \frac{9000}{10} = 900 \, \text{W}
\]
### Step 6: Compare with Given Average Power
We know from the problem statement that:
\[
P_{\text{avg}} = 100,000 \, \text{W}
\]
Setting our calculated average power equal to the given average power:
\[
900 = 100K
\]
Solving for \( K \):
\[
K = \frac{900}{100} = 9
\]
### Final Answer
The value of \( K \) is \( \boxed{9} \).
