A body of mass 1 kg begins to move under the action of a time dependent force `F=(2that(i)+3t^(2)hat(j))N, "where" hat(i)and hat(j)` are unit vector along x and y axis. What power will be developed by the force at the time?

Options

A. `(2t^(3)+3t^(4)) W`
B. `(2t^(3)+3t^(5)) W`
C. `(2t^(2)+3t^(2)) W`
D. `(2t^(2)+3t^(4)) W`

Answer

To solve the problem step by step, we will follow the outlined approach to find the power developed by the force at a given time.

### Step 1: Identify the Given Information
We have:
- Mass of the body, \( m = 1 \, \text{kg} \)
- Force acting on the body, \( \mathbf{F} = (2t \hat{i} + 3t^2 \hat{j}) \, \text{N} \)

### Step 2: Relate Force to Acceleration
Using Newton's second law, we know that:
\[ \mathbf{F} = m \mathbf{a} \]
Since \( m = 1 \, \text{kg} \), we can write:
\[ \mathbf{a} = \frac{\mathbf{F}}{m} = \mathbf{F} = (2t \hat{i} + 3t^2 \hat{j}) \, \text{m/s}^2 \]

### Step 3: Find the Velocity Vector
The acceleration vector \( \mathbf{a} \) is the derivative of the velocity vector \( \mathbf{v} \) with respect to time \( t \):
\[ \mathbf{a} = \frac{d\mathbf{v}}{dt} \]
To find \( \mathbf{v} \), we integrate \( \mathbf{a} \):
\[ \mathbf{v} = \int \mathbf{a} \, dt = \int (2t \hat{i} + 3t^2 \hat{j}) \, dt \]
Calculating the integral:
\[ \mathbf{v} = \left( \int 2t \, dt \right) \hat{i} + \left( \int 3t^2 \, dt \right) \hat{j} = (t^2 + C_1) \hat{i} + (t^3 + C_2) \hat{j} \]
Assuming the initial conditions yield \( C_1 = 0 \) and \( C_2 = 0 \):
\[ \mathbf{v} = t^2 \hat{i} + t^3 \hat{j} \, \text{m/s} \]

### Step 4: Calculate Power Developed by the Force
Power \( P \) developed by the force is given by the dot product of the force and the velocity:
\[ P = \mathbf{F} \cdot \mathbf{v} \]
Substituting the expressions for \( \mathbf{F} \) and \( \mathbf{v} \):
\[ P = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j}) \]
Calculating the dot product:
\[ P = (2t \cdot t^2) + (3t^2 \cdot t^3) = 2t^3 + 3t^5 \]

### Final Expression for Power
Thus, the power developed by the force at time \( t \) is:
\[ P(t) = 2t^3 + 3t^5 \, \text{W} \]