Potential energy (V) versus distance (x) is given by the graph, Rank various regions as per the magnitudes of the force…

Q: Potential energy (V) versus distance (x) is given by the graph, Rank various regions as per the magnitudes of the force (F) acting on a particle from high to low.

### Related Formula ### Core Logic The magnitude of the force is given by the absolute value of the slope of the potential energy versus position curve: |F| = |slope of V-x graph|. Comparing the slopes of different regions from the graph: - Region BC has the steepest slope (largest magnitude). - Region AB has a moderate positive slope. - Region DE has a moderate negative slope, but less steep than AB. - Region CD is perfectly flat (slope = 0). ### Step 1: Ranking Forces Therefore, magnitudes of

Year	2026	2025	2024	Total
Questions	8	15	8	31

More Work, Energy and Power Previous-Year Questions

Q44 jee_main_2026_21_jan_evening Work Done by Variable Force

A body of mass

2 kg

is moving along x-direction such that its displacement as function of time is given by

x(t) = α t² + β t + γ m

, where

α = 1 m/s²

β = 1 m/s

and

γ = 1 m

. The work done on the body during the time interval

t = 2 s

t = 3 s

, is ________ J.

A. $49$ $49$
B. $42$ $42$
C. $24$ $24$
D. $12$ $12$

Solution

Related Formula

W = \int F \cdot dx = F \cdot Δ x (if force is constant)

v = (dx)/(dt), a = (dv)/(dt)

$F = m a$ $F = m a$

Core Logic

Given displacement equation:

x(t) = 1t² + 1t + 1

Velocity is the first derivative: $v(t) = 2t + 1$ $v(t) = 2t + 1$ Acceleration is the second derivative:

a(t) = 2 m/s²

Since acceleration is constant, the force is also constant:

F = m a = 2 \times 2 = 4 N

Step 1: Finding Displacement Interval

We need the displacement during the interval $t=2$ $t=2$ to $t=3$ $t=3$:

x(3) = 3² + 3 + 1 = 9 + 3 + 1 = 13 m

x(2) = 2² + 2 + 1 = 4 + 2 + 1 = 7 m

Total displacement $S = x(3) - x(2) = 13 - 7 = 6 m$

Step 2: Final Conclusion

Work done:

W = F \cdot S = 4 N \times 6 m = 24 J

Pattern Recognition

If $x(t)$ $x(t)$ is a quadratic in $t$ $t$, acceleration is constant. Work done simply becomes $ma \times Δ x$ $ma \times \Delta x$.

Chapter Mix

Class 11 Physics: Kinematics Class 11 Physics: Work, Energy and Power

Q41 jee_main_2026_22_january_evening Conservative Forces and Work Done

Given below are two statements : Statement I: An object moves from position

r₁

to position

r₂

under a conservative force field

F

. The work done by the force is

W = -\int_r₁^r₂ F \cdot dr

. Statement II : Any object moving from one location to another location can follow infinite number of paths. Therefore, the amount of work done by the object changes with the path it follows for a conservative force. In the light of the above statements, choose the correct answer from the options given below :

A. Both Statement I and Statement II are true
B. Statement I is false but Statement II is true
C. Statement I is true but Statement II is false
D. Both Statement I and Statement II are false

Solution

Related Formula

W = \intr₁r₂ F \cdot d r

\oint Fcons \cdot d r = 0

Core Logic

Evaluating Statement I: Work done by a force is defined as $W = +\intr₁r₂ F \cdot d r$ . The negative sign represents potential energy change ( $Δ U = -W$ $\Delta U = -W$), not work done itself. Thus Statement I is incorrect.

Evaluating Statement II: For a conservative force, work done depends ONLY on initial and final positions, independent of the path taken. Thus Statement II is incorrect.

Step 1: Final Conclusion

Both Statement I and Statement II are false.

Pattern Recognition

Conservative force definition: Work done is path-independent ( $W$ $W$ depends only on end points). Formula for work is $+\int F \cdot d r$ $+\int \vec{F} \cdot d\vec{r}$.

Chapter Mix

Class 11 Physics: Work, Energy and Power

Q27 jee_main_2026_23_january_morning Vertical Circular Motion

A small both A of mass m is attached to a massless rigid rod of length 1m pivoted at point P and kept at an angle of

60°

with vertical as shown in figure. At distance of 1m below point P, an identical bob B is kept at rest on a smooth horizontal surface that extends to a circular track of radius R as shown in figure. If bob B just manages to complete the circular path of radius R upto a point Q after being hit elastically by bob A, then radius R is

Vertical Circular Motion diagram for Q27 - JEE Main 2026 Morning — Displays bob A on a rod hitting bob B which enters a circular track.

A. $(3)/(5)$ $\frac{3}{5}$
B. $(1)/(5)$ $\frac{1}{5}$
C. $2+\sqrt(3)5$ $\frac{2+\sqrt{3}}{5}$
D. $2-\sqrt(3)5$ $\frac{2-\sqrt{3}}{5}$

Solution

Related Formula

V = \sqrt(2gl(1 - θ))

Vmin = \sqrt(5gR)

Core Logic

Bob A gains kinetic energy converting from potential energy when it falls from angle $60°$ $60^{\circ}$. Since the collision is elastic and masses are identical, Bob A will transfer its entire momentum to Bob B. Bob B will then use this velocity to complete the vertical circular loop.

Step 1: Velocity of Bob A before collision

Velocity of Bob A at the lowest point:

VA = \sqrt(2gl(1 - θ))

VA = \sqrt(2 \times 10 \times 1 (1 - (1)/(2))) = \sqrt(10) m/s

Step 2: Velocity of Bob B after collision

Due to elastic collision between identical masses, velocities are exchanged.

VB = VA = \sqrt(10) m/s

Step 3: Condition for completing loop

To just complete the circular track of radius $R$ $R$, the minimum velocity required at the lowest point is:

VB = \sqrt(5gR)

\sqrt(10) = \sqrt(5 \times 10 \times R)

$10 = 50R$ $10 = 50R$

R = (1)/(5) m

Pattern Recognition

Sees: "elastic collision" + "identical masses" → velocities swap instantly. "just completes circular path" → velocity is exactly $\sqrt(5gR)$ $\sqrt{5gR}$ at the bottom.

Chapter Mix

Class 11 Physics: Work, Energy and Power Class 11 Physics: Systems of Particles and Rotational Motion

Q32 jee_main_2026_23_january_morning Potential Energy of a Spring

Two blocks with masses 100g and 200g are attached to the ends of springs A and B as shown in figure. The energy stored in A is E. The energy stored in B, when spring constants

kA

kB

of A and B, respectively satisfy the relation

4kA = 3kB

, is :

Potential Energy of a Spring diagram for Q32 - JEE Main 2026 Morning — Two identical-looking springs vertically suspended with masses attached.

A. $4E$ $4E$
B. $2E$ $2E$
C. $3E$ $3E$
D. $(4)/(3)E$ $\frac{4}{3}E$

Solution

Related Formula

$kx = mg$ $kx = mg$

U = (1)/(2)kx² = (1)/(2) m²g²k

Core Logic

For a block suspended in equilibrium, the restoring force of the spring balances gravity. Thus, the extension is $x = mg/k$ $x = mg/k$. Substituting this into the spring energy equation gives energy directly in terms of mass and spring constant.

Step 1: Compare Energy Ratios

Given energy proportional relation:

U \propto m²k

UAUB = ( mAmB)²( kBkA)

Step 2: Insert values

Given $4kA = 3kB kBkA = (4)/(3)$ Also $mA = 100g$ $m_{A} = 100\text{g}$, $mB = 200g mAmB = (1)/(2)$

UAUB = ((1)/(2))²((4)/(3)) = (1)/(4) \times (4)/(3) = (1)/(3)

Step 3: Final Calculation

Since $UA = E$ $U_{A} = E$, we have:

EUB = (1)/(3) UB = 3E

Pattern Recognition

Sees: "energy stored" + "spring under gravity weight" → jump straight to $U = ((mg)²)/(2k)$ $U = \frac{(mg)^2}{2k}$ instead of finding $x$ $x$ first.

Chapter Mix

Class 11 Physics: Work, Energy and Power Class 11 Physics: Laws of Motion

Potential energy (V) versus distance (x) is given by the graph, Rank various regions as per the magnitudes of the force (F) acting on a particle from high to low. The image shows a graph of potential energy versus distance with marked regions AB, BC, CD, and DE.

Solution & Explanation

Related Formula

Core Logic

Step 1: Ranking Forces

Pattern Recognition

Chapter Mix

More Work, Energy and Power Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Finding Displacement Interval

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Velocity of Bob A before collision

Step 2: Velocity of Bob B after collision

Step 3: Condition for completing loop

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compare Energy Ratios

Step 2: Insert values

Step 3: Final Calculation

Pattern Recognition

Chapter Mix

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Potential energy (V) versus distance (x) is given by the graph, Rank various regions as per the magnitudes of the force (F) acting on a particle from high to low.
The image shows a graph of potential energy versus distance with marked regions AB, BC, CD, and DE.