Animated Solution for Physics - Laws of Motion: A body of mass $2\text{ kg}$ moves under a force of $(2\hat{i}+3\hat{j}+5\hat{k})\text{ N}$. It starts from rest and was at the origin initially. After $4\text{ s}$, its new coordinates are $(8,b,20)$. The value of $b$ is ......... . (Round off to the nearest integer)

When Force Meets a Particle at Rest

Imagine a vast, three-dimensional space. A particle with a mass of $2\text{ kg}$ is sitting quietly at the origin $(0,0,0)$. It is completely at rest, waiting for something to disturb its equilibrium. Suddenly, a constant force kicks in, pulling it across the $x$, $y$, and $z$ axes simultaneously. How do we track its journey? Let's break it down step by step.

Newton's Second Law

To find out how this force changes the particle's motion, we call upon Newton's second law of motion. Acceleration is simply the net force divided by the mass:

Let's carefully substitute our known values. The force vector is given as $\vec{F}=2\hat{i}+3\hat{j}+5\hat{k}\text{ N}$, and the mass is $m=2\text{ kg}$.

By dividing each component by two, we get our acceleration vector. Notice how the acceleration points in the exact same direction as the force. This is a fundamental property of Newtonian mechanics.

Kinematics in 3D

Now, we need to find where the particle ends up after $4\text{ seconds}$. Since the force is constant, the acceleration is also constant. This means we can confidently use the vector form of the second equation of motion:

Because the particle started from rest, the initial velocity vector $\vec{u}$ is the zero vector ($\vec{0}$). This makes our calculation much simpler, as the first term vanishes completely. We just plug in our acceleration vector and set the time $t=4\text{ s}$:

Squaring the time gives us $16$. The denominator has a $2$ from the formula and a $2$ from the mass, making it $4$. Dividing $16$ by $4$ gives us a scalar multiplier of $4$:

Distributing this scalar into our vector, we find the final displacement vector:

Comparing Coordinates

Since the particle started at the origin, its final position vector is exactly equal to its displacement vector. Therefore, its final coordinates are $(8,12,20)$.

The problem states that the final coordinates are $(8,b,20)$. By directly comparing our calculated $y$-component with the given coordinates, we can clearly see the missing piece of the puzzle.

We solved this easily because the initial velocity was zero, making the path a straight line. But what if it had an initial velocity in a different direction? The path would curve into a parabola! Keep that in mind for tougher problems.