Question

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 $E_{mech} = 8$ J, the incorrect statement for this system is :

[ where K.E. = kinetic energy ] 
 

• A. at $x = x_2$, K.E. is greatest and the particle is moving at the fastest speed.
• B. at $x<x_1$, K.E. is smallest and the particle is moving at the slowest speed.
• C. at $x>x_4$, K.E. is constant throughout the region.
• D. at $x = x_3$, K.E. = 4 J.

Hint:

In potential energy diagrams, a particle's motion is confined to regions where its total mechanical energy is greater than or equal to the potential energy ($E_{mech} \ge U(x)$). Regions where $U(x)>E_{mech}$ are classically forbidden.

Answer 1

The Correct Option is B

The total mechanical energy of the system is constant, $E_{mech} = K.E. + U(x) = 8 \text{ J}$.
The kinetic energy is given by $K.E. = E_{mech} - U(x) = 8 - U(x)$.
Since K.E. must be non-negative ($K.E. = \frac{1}{2}mv^2 \ge 0$), the particle can only exist in regions where $U(x) \le E_{mech}$.
Let's analyze each statement:
(A) At $x = x_2$, the potential energy $U(x_2)$ is at its minimum value (close to 0 J from the graph).
Therefore, $K.E. = 8 - U(x_2)$ will be maximum. Since $K.E. = \frac{1}{2}mv^2$, maximum K.E. implies the fastest speed. This statement is correct.
(C) For the region $x>x_4$, the graph shows that the potential energy $U(x)$ is constant at a value of 6 J.
So, $K.E. = 8 - 6 = 2 \text{ J}$. Since U(x) is constant, K.E. is also constant. This statement is correct.
(D) At $x = x_3$, the graph shows the potential energy $U(x_3)$ is 4 J.
So, $K.E. = 8 - U(x_3) = 8 - 4 = 4 \text{ J}$. This statement is correct.
(B) For the region $x<x_1$, the graph shows that the potential energy $U(x)$ is constant at 8 J.
According to the principle of conservation of energy, a particle with total energy $E_{mech} = 8$ J can only access regions where $U(x) \le 8$ J.
The region $x<x_1$ has $U(x) = 8$ J, so K.E. would be $8-8=0$. The particle can reach the point $x=x_1$ (this is a turning point), but it cannot enter the region $x<x_1$.
This region is classically forbidden for the particle. Therefore, any statement describing the particle's motion "at $x<x_1$" is fundamentally incorrect because the particle cannot be there.