We are given:
Step 1: Use Newton’s Second Law to find acceleration:
\[ \vec{a} = \frac{\vec{F}}{m} = 2t \hat{i} + 3t^2 \hat{j} \]
Step 2: Integrate to find velocity:
\[ \vec{v} = \int \vec{a} \, dt = \int (2t \hat{i} + 3t^2 \hat{j}) \, dt = t^2 \hat{i} + t^3 \hat{j} \]
Step 3: Power is given by dot product \( \vec{F} \cdot \vec{v} \):
\[ P = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j}) = 2t^3 + 3t^5 \] \[ \Rightarrow P = 2t^3 + 3t^5 \, \text{W} \]
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).