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Work, Energy and Power PYQs · MHT CET · JEE Main · NEET

Potential Energy & Equilibrium

Questions
16
MHT CET
0
JEE Main
11
NEET
5
Years
1995–2026
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Q1JEE Main 2026 · 22 JAN · SHIFT 2Easy

Potential Energy & Equilibrium · Conservative & non-conservative forces

Given below are two statements:

Statement I: An object moves from position r1r_1 to position r2r_2 under a conservative force field F⃗\vec{F}. The work done by the force is W=−∫r1r2F⃗⋅dr→W = -\int_{r_1}^{r_2} \vec{F} \cdot \overrightarrow{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:

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Answer: (c)

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Q2JEE Main 2026 · 21 JAN · SHIFT 1Medium

Potential Energy & Equilibrium · U-x graphs & allowed region

Potential energy ( VV) versus distance ( xx) is given by the graph. Rank various regions as per the magnitudes of the force ( FF) acting on a particle from high to low.

Figure

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Answer: (a)

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Q3JEE Main 2025 · 7 APR · SHIFT 2Easy

Potential Energy & Equilibrium · Conservative & non-conservative forces

Which one of the following forces cannot be expressed in terms of potential energy?

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Answer: (a)

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Q4JEE Main 2025 · 28 JAN · SHIFT 1Medium

Potential Energy & Equilibrium · Conservative & non-conservative forces

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R\mathbf{R}

Assertion A: In a central force field, the work done is independent of the path chosen. Reason R: Every force encountered in mechanics does not have an associated potential energy. In the light of the above statements, choose the most appropriate answer from the options given below

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Answer: (b)

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Q5JEE Main 2024 · 29 JAN · SHIFT 1Medium

Potential Energy & Equilibrium · Force from potential: F = −dU/dx

The potential energy function (in JJ) of a particle in a region of space is given as U=(2x2+3y3+2z)U = \left(2x^2 + 3y^3 + 2z\right). Here x,yx, y and zz are in meter. The magnitude of xx-component of force (in NN) acting on the particle at point P(1,2,3) mP(1, 2, 3)\text{ m} is:

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Answer: (a)

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Q6JEE Main 2022 · 24 JUN · SHIFT 2Medium

Potential Energy & Equilibrium · Equilibrium from U(r)

Potential energy as a function of r is given by U=Ar10−Br5U = \frac{A}{r^{10}} - \frac{B}{r^5}, where r is the interatomic distance, A and B are positive constants. The equilibrium distance between the two atoms will be:

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Answer: (c)

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Q7JEE Main 2021 · 27 JUL · SHIFT 2Medium

Potential Energy & Equilibrium · U-x graphs & allowed region

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 Emech = 8 J, the incorrect statement for this system is:

Figure

[ where K.E. = kinetic energy ]

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Answer: (b)

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Q8JEE Main 2021 · 25 FEB · SHIFT 1Medium

Potential Energy & Equilibrium · Equilibrium from U(r)

The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as

U=αr10−βr5−3U = \frac{\alpha}{r^{10}} - \frac{\beta}{r^5} - 3

where, α\alpha and β\beta are positive constants. The equilibrium distance between two atoms will be (2αβ)ab\left(\frac{2\alpha}{\beta}\right)^{\frac{a}{b}}, where a = ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 1

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Q9JEE Main 2020 · 6 SEP · SHIFT 1Medium

Potential Energy & Equilibrium · Equilibrium from U(r)

If the potential energy between two molecules is given by

U = −Ar6+Br12-\frac{A}{r^6} + \frac{B}{r^{12}},

then at equilibrium, separation between molecules, and the potential energy are:

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Answer: (a)

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Q10NEET 2013Easy

Potential Energy & Equilibrium · U-x graphs & allowed region

A particle with total energy E is moving in a potential energy region U(x). Motion of the particle is restricted to the region when

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Answer: (c)

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Q11NEET 2012 · AIPMT · PRELIMSMedium

Potential Energy & Equilibrium · Equilibrium from U(r)

The potential energy of a particle in a force field is U=Ar2−BrU = \frac{A}{r^2} - \frac{B}{r} where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is

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Answer: (b)

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Q12NEET 2011 · AIPMT · PRELIMSEasy

Potential Energy & Equilibrium · Conservative & non-conservative forces

The potential energy of a system increases if work is done

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Answer: (b)

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Q13NEET 2011 · AIPMT · PRELIMSMedium

Potential Energy & Equilibrium · U-x graphs & allowed region

A particle of mass m is released from rest and follows a parabolic path as shown. Assuming that the displacement of the mass from the origin is small, which graph correctly depicts the position of the particle as a function of time?

Figure

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Answer: (b)

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Q14JEE Main 2010 · AIEEEDifficult

Potential Energy & Equilibrium · Equilibrium from U(r)

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by U(x)=ax12−bx6,U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}, where aa and bb are constants and xx is the distance between the atoms. If the dissociation energy of the molecule is

D=[U(x=∞)−Uat equilibrium],DD = \left[U(x = \infty) - U_{at\ equilibrium}\right], \quad D is

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Answer: (c)

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Q15JEE Main 2006 · AIEEEMedium

Potential Energy & Equilibrium · U-x graphs & allowed region

The potential energy of a 1 kg1\text{ kg} particle free to move along the xx-axis is given by V(x)=(x44−x22) JV(x) = \left(\frac{x^4}{4} - \frac{x^2}{2}\right)\text{ J}. The total mechanical energy of the particle is 2 J.2\text{ J}. Then, the maximum speed (in m/s\text{m/s}) is

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Answer: (a)

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Q16NEET 1995 · AIPMTMedium

Potential Energy & Equilibrium · Equilibrium from U(r)

The potential energy between two atoms, in a molecule, is given by U(x)=ax12−bx6U(x) = \frac{a}{x^{12}} - \frac{b}{x^6} where aa and bb are positive constants and xx is the distance between the atoms. The atom is in stable equilibrium, when

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Answer: (a)

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