A particle of mass `2kg` starts to move at position `x=0` and time `t=0` under the action of force `F=(10+4x)N` along the `x- ` axis on a frictionless horizontal track . Find the power delivered by the force in watts at the instant the particle has moved by distance `5m`.

Explanation

To find the power delivered by the force at the instant the particle has moved by a distance of 5 meters, we can follow these steps:

### Step 1: Identify the force acting on the particle
The force acting on the particle is given by:
\[ F = 10 + 4x \, \text{N} \]

### Step 2: Calculate the work done by the force
The work done \( W \) can be calculated using the formula:
\[ W = \int F \, dx \]
We need to evaluate this integral from \( x = 0 \) to \( x = 5 \):
\[ W = \int_0^5 (10 + 4x) \, dx \]

### Step 3: Solve the integral
Calculating the integral:
\[ W = \int_0^5 (10 + 4x) \, dx = \left[ 10x + 2x^2 \right]_0^5 \]
Now substituting the limits:
\[ W = \left( 10(5) + 2(5^2) \right) - \left( 10(0) + 2(0^2) \right) = 50 + 50 = 100 \, \text{J} \]

### Step 4: Relate work done to kinetic energy
According to the work-energy theorem:
\[ W = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]
Since the particle starts from rest, \( KE_{\text{initial}} = 0 \):
\[ W = KE_{\text{final}} = \frac{1}{2} mv^2 \]
Substituting the values:
\[ 100 = \frac{1}{2} (2)v^2 \]
This simplifies to:
\[ 100 = v^2 \implies v = \sqrt{100} = 10 \, \text{m/s} \]

### Step 5: Calculate the force at \( x = 5 \)
Now we can calculate the force when \( x = 5 \):
\[ F = 10 + 4(5) = 10 + 20 = 30 \, \text{N} \]

### Step 6: Calculate the power delivered by the force
Power \( P \) is given by:
\[ P = F \cdot v \]
Substituting the values we found:
\[ P = 30 \, \text{N} \cdot 10 \, \text{m/s} = 300 \, \text{W} \]

### Final Answer:
The power delivered by the force at the instant the particle has moved by a distance of 5 meters is:
\[ \boxed{300 \, \text{W}} \]