Question

A body of mass \(2 \, \text{kg}\) begins to move under the action of a time-dependent force given by \(\vec{F} = (6t \, \hat{i} + 6t^2 \, \hat{j}) \, \text{N}\). The power developed by the force at the time \(t\) is given by:

• A. \((6t^4 + 9t^5) \, \text{W}\)
• B. \((3t^3 + 6t^5) \, \text{W}\)
• C. \((9t^5 + 6t^3) \, \text{W}\)
• D. \((9t^3 + 6t^5) \, \text{W}\)

Answer 1

The Correct Option is D

To find the power developed by a time-dependent force acting on a body, we follow these steps:

1. The force acting on the body is given by the vector function: 

\[\vec{F} = (6t \, \hat{i} + 6t^2 \, \hat{j}) \, \text{N}\]

1. .
2. The velocity of the body, which is initially at rest, can be found by integrating the acceleration. The acceleration is given by: 

\[\vec{a} = \frac{\vec{F}}{m} = \left(\frac{6t}{2} \, \hat{i} + \frac{6t^2}{2} \, \hat{j}\right) \, \text{m/s}^2 = (3t \, \hat{i} + 3t^2 \, \hat{j}) \, \text{m/s}^2\]

1. .
2. Integrating the acceleration with respect to time to find the velocity: 

\[\vec{v} = \int \vec{a} \, dt = \int (3t \, \hat{i} + 3t^2 \, \hat{j}) \, dt\]

1. .

\[\vec{v} = (3 \frac{t^2}{2} \, \hat{i} + 3 \frac{t^3}{3} \, \hat{j}) = \left(\frac{3t^2}{2} \, \hat{i} + t^3 \, \hat{j}\right)\]

• .

1. Now, compute the power developed by the force, which is given by: 

\[\text{Power} = \vec{F} \cdot \vec{v}\]

1.  (dot product of force and velocity vectors).
   • Substitute \(\vec{F} = (6t \, \hat{i} + 6t^2 \, \hat{j})\) and \(\vec{v} = \left(\frac{3t^2}{2} \, \hat{i} + t^3 \, \hat{j}\right)\):

\[\vec{F} \cdot \vec{v} = (6t) \left(\frac{3t^2}{2}\right) + (6t^2) (t^3)\]

• Simplifying gives:

\[\vec{F} \cdot \vec{v} = 9t^3 + 6t^5\]

1. Thus, the expression for the power developed by the force at any time \(t\) is: 

\[(9t^3 + 6t^5) \, \text{W}\]

1. .

Therefore, the correct option is: (9t³ + 6t⁵) W.

Answer 2

Given:

\[ \vec{F} = (6t \, \hat{i} + 6t^2 \, \hat{j}) \, \text{N} \]

The mass of the body is \( m = 2 \, \text{kg} \). According to Newton's second law:

\[ \vec{F} = m\vec{a} \implies \vec{a} = \frac{\vec{F}}{m} = \left(3t \, \hat{i} + 3t^2 \, \hat{j}\right) \, \text{m/s}^2 \]

The velocity \(\vec{v}\) is obtained by integrating the acceleration:

\[ \vec{v} = \int \vec{a} \, dt = \int \left(3t \, \hat{i} + 3t^2 \, \hat{j}\right) dt = \left(\frac{3t^2}{2} \, \hat{i} + t^3 \, \hat{j}\right) \, \text{m/s} \]

The power developed by the force is given by:

\[ P = \vec{F} \cdot \vec{v} \]

Calculating the dot product:

\[ P = (6t \, \hat{i} + 6t^2 \, \hat{j}) \cdot \left(\frac{3t^2}{2} \, \hat{i} + t^3 \, \hat{j}\right) \]

\[ P = 6t \cdot \frac{3t^2}{2} + 6t^2 \cdot t^3 \]

\[ P = 9t^3 + 6t^5 \, \text{W} \]