A body of mass 2 kg moving with velocity of $ \vec{v}_{in} = 3 \hat{i} + 4 \hat{j} \text{ ms}^{-1} $ enters into a constant force field of 6N directed along positive z-axis. If the body remains in the field for a period of $ \frac{5}{3} $ seconds, then velocity of the body when it emerges from force field is.

<p>To determine the velocity of the body as it emerges from the force field, we can break this process down as follows:</p> <p><p><strong>Force and Acceleration Calculation:</strong></p> <p>The body is subjected to a constant force of 6 N, which acts along the positive z-axis. The mass of the body is 2 kg.</p> <p>Using Newton's second law, $ \vec{F} = m \vec{a} $, the acceleration $ \vec{a} $ can be calculated as:</p> <p>$ \vec{a} = \frac{\vec{F}}{m} = \frac{6 \hat{k}}{2} = 3 \hat{k} \text{ ms}^{-2} $</p></p> <p><p><strong>Initial Velocity:</strong></p> <p>The initial velocity of the body is given by:</p> <p>$ \vec{u} = 3 \hat{i} + 4 \hat{j} \text{ ms}^{-1} $</p></p> <p><p><strong>Time of Motion:</strong></p> <p>The time the body spends in the force field is:</p> <p>$ t = \frac{5}{3} \text{ seconds} $</p></p> <p><p><strong>Final Velocity Calculation:</strong></p> <p>The final velocity $ \vec{v} $ is given by the equation of motion:</p> <p>$ \vec{v} = \vec{u} + \vec{a} t $</p> <p>Substituting the known values:</p> <p>$ \vec{v} = (3 \hat{i} + 4 \hat{j}) + 3 \hat{k} \left(\frac{5}{3}\right) $</p> <p>Simplifying the expression gives:</p> <p>$ \vec{v} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} $</p></p> <p>Therefore, the velocity of the body as it exits the force field is $ 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \text{ ms}^{-1} $.</p>