Animated Solution for Physics - Work, Energy, and Power: Given below is the plot of a potential energy function $U(x)$ for a system, in which a particle is in one-dimensional motion, while a conservative force $F(x)$ acts on it. Suppose that $E_{\text{mech}}=8\text{ J}$, the incorrect statement for this system is [where, $\text{KE}=\text{kinetic energy}$]

Imagine you are looking at a roller coaster track. The potential energy graph $U(x)$ tells us exactly how the "height" of this track changes as our particle moves along the $x$-axis. We are given a crucial piece of information: the total mechanical energy of the system, $E_{\text{mech}}$, is fixed at $8\text{ J}$.

This total energy acts as a strict budget. The particle can trade potential energy for kinetic energy, but it can never spend more than $8\text{ J}$ in total.

The core principle governing this system is the conservation of mechanical energy. The total energy is simply the sum of kinetic energy ($K$) and potential energy ($U$):

By rearranging this, we get our master equation for kinetic energy:

This simple subtraction is the key to unlocking the entire problem. Let's use it to test each option.

Testing Option (a): Look at the far right of the graph, where $x>x_4$. The potential energy curve is perfectly flat at $U=6\text{ J}$. Plugging this into our master equation, we get $K=8-6=2\text{ J}$. Since the potential energy doesn't change in this region, the kinetic energy remains a constant $2\text{ J}$. Option (a) is a true statement.

Testing Option (c): Now, let's find the lowest point on the curve, which occurs at $x=x_2$. Here, the potential energy hits rock bottom at $U=0\text{ J}$. This means the kinetic energy must be at its absolute maximum: $K=8-0=8\text{ J}$. Since kinetic energy is directly proportional to the square of the speed ($K=\frac{1}{2}m{v}^2$), maximum kinetic energy guarantees the fastest speed. Option (c) is also true.

Testing Option (d): At the specific point $x=x_3$, the graph shows the potential energy is exactly $U=4\text{ J}$. Subtracting this from our $8\text{ J}$ budget leaves us with $K=4\text{ J}$. Option (d) is perfectly accurate.

Testing Option (b): Finally, let's examine the region on the far left, where $x<x_1$. The graph shows the potential energy is $U=8\text{ J}$. This consumes our entire energy budget!

The kinetic energy is indeed at its smallest possible value (zero). However, there is a massive catch. If the kinetic energy is zero, the velocity must also be zero. The particle is completely at rest.

Option (b) claims the particle is "moving at the slowest speed." While zero is technically the lowest number for speed, a particle with zero speed is not moving at all. This subtle semantic trick makes the statement technically incorrect. Therefore, option (b) is the false statement we were looking for!