1. Physics
  2. Work, Energy and Power
  3. Potential Energy

Work, Energy and Power · JEE & NEET Physics

Potential Energy: notes and previous year questions

Energy stored by position or shape: mgh, ½kx², W = −ΔU, F = −dU/dx, equilibrium, and reading a U–x graph.

Potential Energy in short

  • Potential energy is energy stored by position or shape, only for conservative forces.
  • We choose where U is zero; only changes in U have meaning.
  • Near the Earth U = mgh; far away U = −GMm/r, negative because the body is bound.
  • A spring stores ½kx², the area of its force–stretch triangle, the same for stretch and squeeze.

1Stored energy

Lift a 2 kg ball slowly up 5 m: you do mgh=100mgh = 100 J of work, but the ball does not speed up. The energy is stored. Let the ball fall and the same 100 J comes back as kinetic energy.

Potential energy is energy stored in a system because of the position or the shape (configuration) of its parts.

  • It exists only for conservative forces: gravity, spring forces, electric forces.
  • It can be positive, negative or zero, because we choose where it is zero.
  • Only changes in potential energy have physical meaning.

2Gravitational potential energy

Ug=mghU_g = mghNear the Earth's surface; h is the height above the level you choose as zero.

Choose any handy zero: the ground, the starting point, or the lowest point in the problem. Raising the body by Δh\Delta h always changes UU by mgΔhmg\Delta h.

Far from the Earth, gravity weakens and the full formula takes over, with zero at infinite distance:

U=−GMmrU = -\frac{GMm}{r}Negative everywhere: the body is bound, and energy must be added to pull it away.

3Spring potential energy

To stretch a spring by xx, you pull with a force that grows from 0 to kxkx. The work is the area of that triangle on the force–stretch graph, 12×x×kx\tfrac{1}{2} \times x \times kx, and it is stored in the spring.

Us=12kx2U_s = \tfrac{1}{2}kx^2k in N/m; x is the stretch or the squeeze from natural length.
  • Never negative, and the same for a stretch and an equal squeeze.
  • Zero at natural length, largest when the spring is most deformed.

4Work and potential energy

Throw a ball up: gravity opposes the motion and does negative work while the potential energy rises. On the way down, gravity does positive work and the potential energy falls by the same amount.

Wconservative=−ΔU=Ui−UfW_{\text{conservative}} = -\Delta U = U_i - U_fThis rule is what defines potential energy.

5Force from potential energy

For a small step dxdx, the work is F dx=−dUF\,dx = -dU. So the force is minus the slope of the potential energy:

F=−dUdxF = -\frac{dU}{dx}
F⃗=−(∂U∂xi^+∂U∂yj^+∂U∂zk^)\vec F = -\left(\frac{\partial U}{\partial x}\hat i + \frac{\partial U}{\partial y}\hat j + \frac{\partial U}{\partial z}\hat k\right)In three dimensions.

Picture UU as a landscape: the force always points downhill, toward lower potential energy, and it is strongest where the graph is steepest.

  • Gravity: U=mghU = mgh gives F=−mgF = -mg, straight down.
  • Spring: U=12kx2U = \tfrac{1}{2}kx^2 gives F=−kxF = -kx, back toward natural length.
  • U=3x2+5x+2U = 3x^2 + 5x + 2 J gives F=−(6x+5)F = -(6x + 5): −11-11 N at x=1x = 1 m and −17-17 N at x=2x = 2 m.

6Equilibrium

Where dU/dx=0dU/dx = 0 the force is zero: the body can rest there, in equilibrium. Whether it stays after a nudge depends on the shape of the curve.

TypeU curve$d^2U/dx^2$After a nudge
stableminimum (a valley)>0> 0comes back
unstablemaximum (a hilltop)<0< 0moves away
neutralflat=0= 0stays at its new place

7Reading a U–x graph

Draw the total energy EE as a flat line on the UU–xx graph. At every point the kinetic energy is the gap between the line and the curve:

KE=E−U≥0KE = E - U \ge 0
  • The body can move only where U≤EU \le E.
  • It turns back at the turning points, where U=EU = E and KE=0KE = 0.
  • With more energy the range widens, and the body may cross a hilltop into the next valley.

For U=x4−4x2U = x^4 - 4x^2 with E=0E = 0, a body in the right-hand valley moves between x=0x = 0 and x=2x = 2 m. With E=2E = 2 J it rolls over the hilltop at x=0x = 0 (where U=0U = 0) and visits both valleys, but it can never escape, because far out UU rises above 2 J.

8Common potential energies

SystemPotential energyForce
gravity near the Earthmghmghmgmg, downward
spring12kx2\tfrac{1}{2}kx^2−kx-kx, back to natural length
gravity, general−GMm/r-GMm/rGMm/r2GMm/r^2, attractive
two chargeskq1q2/rkq_1q_2/rkq1q2/r2kq_1q_2/r^2, along the line joining them

Summary

Key ideas

  • Potential energy is energy stored by position or shape, only for conservative forces.
  • We choose where U is zero; only changes in U have meaning.
  • Near the Earth U = mgh; far away U = −GMm/r, negative because the body is bound.
  • A spring stores ½kx², the area of its force–stretch triangle, the same for stretch and squeeze.
  • The work of a conservative force is minus the change in its potential energy.
  • The force is minus the slope of U: it points downhill on the U–x graph.
  • Equilibrium is where dU/dx = 0: stable at a minimum, unstable at a maximum, neutral where flat.
  • With total energy E, the body moves only where U ≤ E and turns back where U = E.

Every equation

Gravity near the Earth
U=mghU = mgh
Gravity, general
U=−GMmrU = -\frac{GMm}{r}
Spring
U=12kx2U = \tfrac{1}{2}kx^2
Work of a conservative force
W=−ΔU=Ui−UfW = -\Delta U = U_i - U_f
Force from U (1D)
F=−dUdxF = -\frac{dU}{dx}
Force from U (3D)
F⃗=−∇U\vec F = -\nabla U
Equilibrium
dUdx=0\frac{dU}{dx} = 0
Stable
d2Udx2>0\frac{d^2U}{dx^2} > 0
Unstable
d2Udx2<0\frac{d^2U}{dx^2} < 0
Kinetic energy from the graph
KE=E−UKE = E - U
Two charges
U=kq1q2rU = \frac{kq_1q_2}{r}

Previous year questions with solutions

Real JEE and NEET questions on potential energy. Try each one before you open the solution.

Q1JEE Main 2026One correct option

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⋅dr⃗W=-{\int}_{r_{1}}^{r_{2}}\vec{F}\cdot \vec{{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 :

  1. AStatement I is true but Statement II is false
  2. BStatement I is false but Statement II is true
  3. CBoth Statement I and Statement II are false
  4. DBoth Statement I and Statement II are true
Show answer and solution

Answer: Option C

Both are false. Statement I has a sign the wrong way round: the work done by the force is W=+∫r1r2F⃗⋅dr⃗W = +\int_{r_{1}}^{r_{2}} \vec{F}\cdot d\vec{r}, the sum of F⃗⋅dr⃗\vec{F}\cdot d\vec{r} over the little steps. With the minus sign in front, that integral is ΔU=U2−U1\Delta U = U_{2} - U_{1}, the change in potential energy, not the work. Statement II gets the one defining property of a conservative force backwards: there are indeed infinitely many paths, but the work along every one of them is the same, U1−U2U_{1} - U_{2}. A force whose work did change with the path — friction — is exactly the kind that has no potential energy.

Q2NEET 2015One correct option

Two similar springs P and Q have spring constants K_{P} and K_{Q}, such that K_{P} > K_{Q}. They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs W_{P} and W_{Q} are related as, in case (a) and case (b) respectively

  1. AWP>WQ;WQ>WPW_{P} > W_{Q}; W_{Q} > W_{P}
  2. BWP<WQ;WQ<WPW_{P} < W_{Q}; W_{Q} < W_{P}
  3. CWP=WQ;WP>WQW_{P} = W_{Q}; W_{P} > W_{Q}
  4. DWP=WQ;WP=WQW_{P} = W_{Q}; W_{P} = W_{Q}
Show answer and solution

Answer: Option A

The work done on each spring is the energy it ends up storing; here it is sizes that are compared. Case (a), same extension: W=12Kx2∝KW = \tfrac{1}{2}Kx^{2} \propto K, and KP>KQK_{P} > K_{Q}, so WP>WQW_{P} > W_{Q}. Case (b), same force: W=F22K∝1KW = \dfrac{F^{2}}{2K} \propto \dfrac{1}{K}, so now the softer spring wins and WQ>WPW_{Q} > W_{P}. The order flips between the two cases, which is the whole point of the question. B has both the wrong way round; C and D assume the stiffness cannot matter in one of the cases.

Q3JEE Main 2026One correct option

Two blocks with masses 100 g and 200 g hang at rest from the lower ends of two vertical springs AA and BB respectively, whose upper ends are fixed to a ceiling. The energy stored in AA is EE. The energy stored in BB, when spring constants kA,kBk_{A},k_{B} of AA and BB, respectively satisfy the relation 4kA=3kB4k_{A}=3k_{B}, is :

  1. A4E4E
  2. B2E2E
  3. C43E\frac{4}{3}E
  4. D3E3E
Show answer and solution

Answer: Option D

Each block hangs at rest, so its spring pulls up with exactly the block's weight, and the energy stored is U=F22k=(mg)22kU = \dfrac{F^{2}}{2k} = \dfrac{(mg)^{2}}{2k}. For AA: E=(mAg)22kAE = \dfrac{(m_{A}g)^{2}}{2k_{A}}. For BB the force is doubled, since mB=2mAm_{B} = 2m_{A}, and the constant is kB=43kAk_{B} = \tfrac{4}{3}k_{A}: EB=4(mAg)22×43kA=3×(mAg)22kA=3EE_{B} = \dfrac{4(m_{A}g)^{2}}{2 \times \tfrac{4}{3}k_{A}} = 3 \times \dfrac{(m_{A}g)^{2}}{2k_{A}} = 3E. Doubling the force alone would give 4E4E, option A; the stiffer spring takes back a factor 34\tfrac{3}{4}. Mind which way 4kA=3kB4k_{A} = 3k_{B} goes: it makes BB the stiffer spring, not the softer.

Practice questions, easy to hard

Three questions from the potential energy practice ladder: one easy, one medium, one hard.

Q4One correct option

A spring's force is conservative too: the work done by the spring as its end goes from extension xix_{i} to xfx_{f} is 12k(xi2−xf2)\tfrac{1}{2}k\left(x_{i}^{2} - x_{f}^{2}\right), which depends on the two ends alone. So a spring has a potential energy, found from Wc=−ΔUW_{c} = -\Delta U with U=0U = 0 at its natural length.

What is the potential energy of a spring of constant kk stretched or compressed by xx?

  1. Akx2kx^{2}
  2. B12kx2\tfrac{1}{2}kx^{2}
  3. C12kx\tfrac{1}{2}kx
  4. D−12kx2-\tfrac{1}{2}kx^{2}
Show answer and solution

Answer: Option B

From the natural length to xx the spring does 12k(0−x2)=−12kx2\tfrac{1}{2}k(0 - x^{2}) = -\tfrac{1}{2}kx^{2}, so ΔU=+12kx2\Delta U = +\tfrac{1}{2}kx^{2}, and with U=0U = 0 at the natural length, U=12kx2U = \tfrac{1}{2}kx^{2}. It is exactly the work you did stretching it, now kept inside the spring. Because xx is squared, a compression stores energy just as a stretch does — squash a spring and it can push something away. D is the spring's own work, sign and all; A drops the half that comes from the triangle under F=kxF = kx; C is not even an energy.

Q5Numerical answer

Equilibrium is where dUdx=0\dfrac{dU}{dx} = 0. For UU with xx in a denominator, the power rule you know still works with negative powers: 1xn=x−n\dfrac{1}{x^{n}} = x^{-n} differentiates to −n x−n−1=−nxn+1-n\,x^{-n-1} = -\dfrac{n}{x^{n+1}}.

A particle moving along x>0x > 0 has U=4x+xU = \dfrac{4}{x} + x (UU in J\mathrm{J}, xx in m\mathrm{m}). At what xx, in m\mathrm{m}, is it in equilibrium?

Show answer and solution

Answer: 2 m

dUdx=−4x2+1\dfrac{dU}{dx} = -\dfrac{4}{x^{2}} + 1, which is zero when x2=4x^{2} = 4, so x=2 mx = 2\ \mathrm{m} (the negative root is outside the region). It is a stable one: close to x=0x = 0 the 4x\dfrac{4}{x} term makes UU enormous, far out the xx term makes it grow again, so x=2 mx = 2\ \mathrm{m} sits at the bottom of a dip, where U=2+2=4 JU = 2 + 2 = 4\ \mathrm{J}. Setting UU itself to zero instead of its slope finds nothing here — UU is never zero for x>0x > 0 — and is the wrong condition anyway.