Question

A body of mass \(1 \, \text{kg}\) begins to move under the action of a time-dependent force \[ \vec{F} = (t \, \hat{i} + 3t^2 \, \hat{j}) \, \text{N}, \] where \(\hat{i}\) and \(\hat{j}\) are unit vectors along \(x\) and \(y\) axes. The power developed by the above force, at the time \(t = 2 \, \text{s}\), will be _____ W.

Hint:

The power developed by a force is \[ P = \vec{F} \cdot \vec{v}. \] Ensure both \(\vec{F}\) and \(\vec{v}\) are evaluated at the same instant of time.

Answer 1

Correct Answer: 100

To find the power developed by the force at \( t = 2 \, \text{s} \), we first need to determine the velocity of the body as a function of time. The given force is \(\vec{F} = (t \, \hat{i} + 3t^2 \hat{j}) \, \text{N}\), and since the mass \( m = 1 \, \text{kg} \), we have the acceleration \(\vec{a} = \vec{F}\).

Starting with the acceleration:
\(\vec{a}(t) = (t \, \hat{i} + 3t^2 \hat{j}) \, \text{m/s}^2\).

Integrating with respect to time to find velocity:
\(\vec{v}(t) = \int \vec{a}(t) \, dt = \int (t \, \hat{i} + 3t^2 \hat{j}) \, dt = \left(\frac{t^2}{2} \, \hat{i} + t^3 \, \hat{j}\right) + \vec{C},\)
where \(\vec{C}\) is the integration constant. Assuming initial velocity \(\vec{v}(0) = 0\), \(\vec{C} = 0\).

Thus, \(\vec{v}(t) = \frac{t^2}{2} \, \hat{i} + t^3 \, \hat{j}\).

The power \( P \) is given by \( P = \vec{F} \cdot \vec{v} \).
Substituting the expressions for \(\vec{F}\) and \(\vec{v}\):
\[\begin{aligned} P(t) &= (t \, \hat{i} + 3t^2 \hat{j}) \cdot \left(\frac{t^2}{2} \, \hat{i} + t^3 \, \hat{j}\right) \\ &= \left(t \cdot \frac{t^2}{2}\right) + \left(3t^2 \cdot t^3\right) \\ &= \frac{t^3}{2} + 3t^5.\end{aligned}\]

Substituting \( t = 2 \, \text{s} \):
\[\begin{aligned} P(2) &= \frac{2^3}{2} + 3 \cdot 2^5 \\ &= \frac{8}{2} + 3 \cdot 32 \\ &= 4 + 96 \\ &= 100 \, \text{W}.\end{aligned}\]

The computed power at \( t = 2 \, \text{s} \) is \( 100 \, \text{W} \), which falls within the given range of 100,100. Hence, the solution is verified.