Shape the potential.

Move the energy level to reveal allowed regions. Watch where the moving particle reverses direction.

Potential family

Quartic coefficient α0.18

Quadratic coefficient β1.35

Total energy E (J)−0.18

Particle mass (kg)1.20

Starting cageStarting direction

Animation speed1.00×

The barrier divides the allowed area into separate wells when E is below zero.

EXAM FOUNDATION

Why does it turn?

For a classical particle subject to a conservative force, the accessible region is determined by total mechanical energy.

JEE / NEET PRACTICE

Concept questions, calculations, and misconception checks with explanations.

Question 1 / 12

CLASSROOM STUDIO

Run targeted predictions and connect the visual model with exam mathematics.

MODEL ACTIVE · CONSERVATIVE

JEE / NEET Physics worksheet · Class 11–12 · Work, Energy and Power

Name: _______________________________ Date: ___________

Interpret U(x), identify classically allowed regions, calculate turning points and speed, and distinguish a velocity reversal from zero force.

Derive the condition for a classically allowed coordinate from energy conservation.
For U(x)=½kx², k=2 N/m, E=4 J, calculate both turning points.
For U(x)=αx⁴−βx² with α=0.18, β=1.35, determine the barrier height and the number of accessible cages when E=−0.18 J.
Explain why v=0 at a turning point does not imply F=0.
A 1.0 kg particle has E=5 J and U(x)=3 J. Find its speed.
For U(x)=−U₀/[1+(x/w)²], identify whether E=0.3 J is a bound state or an escaping one.

From ½mv²+U=E, the kinetic energy ½mv²=E−U≥0; hence E≥U(x).
E=½kx²=x², hence x=±2 m.
Central barrier U(0)=0 J; at E=−0.18 J there are two disconnected allowed intervals.
At a turning point v=0, but F=−dU/dx can be nonzero, causing acceleration back toward the well.
v=√[2(E−U)/m]=√4=2 m/s.
E>0 exceeds the far-field potential limit U→0 from below, so the particle is unbound; it may escape.