Animated Solution for Physics - Work, Energy, and Power: As shown in the figure, a particle of mass $10\text{ kg}$ is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed of the particle at B is $x\text{ m/s}$. (Take, $g=10{\text{ m/s}}^2$) The value of $x$ to the nearest integer is ....... .

Welcome to a classic problem from the realms of Work, Energy, and Power! This question is a beautiful demonstration of one of the most powerful principles in all of physics: the Conservation of Mechanical Energy.

Imagine you are standing at the top of a roller coaster. Before the drop, you have a lot of potential energy but no kinetic energy. As you plunge downwards, that potential energy is magically transformed into kinetic energy, giving you that thrilling rush of speed. This problem is exactly like that roller coaster ride, just scaled down to a particle on a track.

We have a particle of mass $m=10\text{ kg}$ resting at point A. This point is at a height $h_A=10\text{ m}$ above our reference ground level. The particle is slightly nudged, meaning it starts its journey from rest. Therefore, its initial velocity $v_A=0\text{ m/s}$.

It slides down a smooth, curved track to reach point B, which is at a lower height $h_B=5\text{ m}$. We are tasked with finding its speed $v_B$ (which the question calls $x$) at this new point.

Conservation of Energy

The beauty of physics is that we don't need to know the exact mathematical shape of the curved track. Because the track is perfectly smooth, there is no friction to steal our energy and convert it into unrecoverable heat. The only force doing work on the particle is gravity, which is a conservative force.

This means the total mechanical energy of the particle will remain perfectly conserved as it slides down. The sum of its kinetic energy ($K$) and potential energy ($U$) at any point on the track will always be a constant.

Let us translate this physical reality into a mathematical equation. The total energy at point A must equal the total energy at point B:

At point A, the particle is released from rest, so its initial kinetic energy is simply zero ($K_A=0$). Its potential energy is mass times gravity times its height ($U_A=mg{h}_A$).

As it reaches point B, it has picked up some speed $v_B$, giving it a kinetic energy of $K_B=\frac{1}{2}m{v}_B^2$. It is still $5\text{ m}$ above the ground, so it retains some potential energy $U_B=mg{h}_B$.

Equating the two states, we get:

We can group the potential energy terms together to see the energy transformation more clearly:

Notice a profound realization here: the mass '$m$' beautifully cancels out from both sides of the equation!

This tells us that whether the particle is a tiny marble or a massive boulder, it will reach point B with the exact same speed. The mass of the particle does not even matter for this kinematic outcome.

Let us substitute the given values and get the final answer. We know $g=10{\text{ m/s}}^2$, $h_A=10\text{ m}$, and $h_B=5\text{ m}$. The height difference $(h_A-h_B)$ is simply $5\text{ m}$.

Multiplying both sides by 2, we isolate the velocity squared:

Taking the square root, we find that $v_B$ is exactly $10\text{ m/s}$.

Since the question asks for the value of $x$ (where $v_B=x$), our final integer answer is 10.