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

Answer: (D)

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

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:

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

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

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