Question 45: Laws of Motion

Question

A body of mass 2 kg moving with velocity of $ \vec{v}_{in} = 3 \hat{i} + 4 \hat{j} 	ext{ 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.

Accepted Answer

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

Answer

$ 3\hat{i} + 4\hat{j} + \sqrt{5} \hat{k} $

Answer

$ 4\hat{i} + 3\hat{j} + 5\hat{k} $

Answer

$ 3\hat{i} + 4\hat{j} - 5\hat{k} $

Answer

$ 3\hat{i} + 4\hat{j} + 5\hat{k} $

Laws of Motion

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