A body of mass 1 kg begins to move under the action of a time dependent force F=(2t^i+3t2^j)N, where ^i and ^j are unit vectors along x and y axis. What power will be developed by the force at the time t?

JEE Mains - Work, Power & Energy

Question: A body of mass \( m = 1\,\text{kg} \) begins to move under the action of a time-dependent force \[ \vec{F}(t) = t\,\hat{i} + 3t^2\,\hat{j}\,\text{N}, \] where \( \hat{i} \) and \( \hat{j} \) are the unit vectors along the \( x \) and \( y \) axes, respectively. Find the power developed by the force at \( t = 2\,\text{s} \).
[24-Jan-2023 Shift 2]

Select the correct answer:

(A) \(50\,\text{W}\)

(B) \(100\,\text{W}\)

(C) \(150\,\text{W}\)

(D) \(200\,\text{W}\)

Solution

The power developed by a force is given by: \[ P(t) = \vec{F}(t) \cdot \vec{v}(t), \] where \( \vec{v}(t) \) is the velocity vector at time \( t \). Given that the mass \( m = 1\,\text{kg} \), the acceleration is: \[ \vec{a}(t) = \frac{\vec{F}(t)}{m} = t\,\hat{i} + 3t^2\,\hat{j}. \]

Since the body starts from rest (\( \vec{v}(0) = 0 \)), we can find the velocity by integrating the acceleration:

For the \( x \)-component: \[ v_x(t) = \int_0^t \tau\,d\tau = \frac{t^2}{2}. \] For the \( y \)-component: \[ v_y(t) = \int_0^t 3\tau^2\,d\tau = t^3. \] Thus, the velocity vector is: \[ \vec{v}(t) = \frac{t^2}{2}\,\hat{i} + t^3\,\hat{j}. \]

At \( t = 2\,\text{s} \):
\( v_x(2) = \frac{(2)^2}{2} = 2\,\text{m/s} \) and \( v_y(2) = (2)^3 = 8\,\text{m/s} \).

Similarly, the force at \( t = 2\,\text{s} \) is: \[ \vec{F}(2) = 2\,\hat{i} + 3(2)^2\,\hat{j} = 2\,\hat{i} + 12\,\hat{j}\,\text{N}. \]

Now, the power at \( t = 2\,\text{s} \) is: \[ P(2) = \vec{F}(2) \cdot \vec{v}(2) = (2 \times 2) + (12 \times 8) = 4 + 96 = 100\,\text{W}. \] Hence, the power developed by the force at \( t = 2\,\text{s} \) is \( 100\,\text{W} \).