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

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given force
The force acting on the body is given as:
\[ \mathbf{F} = (2t \hat{i} + 3t^2 \hat{j}) \, \text{N} \]

### Step 2: Calculate acceleration
Using Newton's second law, the acceleration \(\mathbf{a}\) can be calculated using the formula:
\[ \mathbf{a} = \frac{\mathbf{F}}{m} \]
where \(m = 1 \, \text{kg}\). Therefore,
\[ \mathbf{a} = \frac{(2t \hat{i} + 3t^2 \hat{j})}{1} = (2t \hat{i} + 3t^2 \hat{j}) \, \text{m/s}^2 \]

### Step 3: Find the velocity vector
The velocity vector \(\mathbf{v}\) can be found by integrating the acceleration with respect to time:
\[ \mathbf{v} = \int \mathbf{a} \, dt = \int (2t \hat{i} + 3t^2 \hat{j}) \, dt \]

Calculating the integrals separately:
1. For the \(\hat{i}\) component:
\[ \int 2t \, dt = t^2 + C_1 \]
2. For the \(\hat{j}\) component:
\[ \int 3t^2 \, dt = t^3 + C_2 \]

Assuming the initial conditions yield \(C_1 = 0\) and \(C_2 = 0\) (the body starts from rest), we have:
\[ \mathbf{v} = (t^2 \hat{i} + t^3 \hat{j}) \, \text{m/s} \]

### Step 4: Calculate power
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 we have:
\[ P = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j}) \]

Calculating the dot product:
\[ P = (2t \cdot t^2) + (3t^2 \cdot t^3) \]
\[ P = 2t^3 + 3t^5 \]

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

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