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Work, Energy & Power revision notes

Every idea in the Work, Energy & Power chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 26 and download the PDF.

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How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Work by constant and variable force

Work by a constant force
7× asked, last 2026
W=F⃗⋅d⃗=Fdcos⁡θ,d⃗=r⃗2−r⃗1W=\vec F\cdot\vec d = Fd\cos\theta,\qquad \vec d=\vec r_2-\vec r_1
F⃗\vec F constant; d⃗\vec d = displacement of the point where the force acts. Several forces: add them first.
mmFFFcos⁡θF\cos\thetaθ\thetadisplacement dd
Only the part of FF along the motion works: W=Fdcos⁡θW=Fd\cos\theta
  • Sign: positive when the force has a component along the motion, zero at 90∘90^\circ, negative against it. Gravity on a bucket being lifted and air drag on a swinging pendulum do negative work.
  • Equal work through the same distance with forces at different angles: Fcos⁡θF\cos\theta must be equal, so F1/F2=cos⁡θ2/cos⁡θ1F_1/F_2=\cos\theta_2/\cos\theta_1.
  • Not distance × direction vector. 'dd along (ai^+bj^)(a\hat i+b\hat j)' means d⃗=d (ai^+bj^)/a2+b2\vec d=d\,(a\hat i+b\hat j)/\sqrt{a^2+b^2}. Make the unit vector first, then dot it with the net force.
Work by gravity
5× asked, last 2024
Wg=−mg Δh=mg (hi−hf),U=mghW_g=-mg\,\Delta h = mg\,(h_i-h_f),\qquad U=mgh
Near Earth (h≪Rh\ll R). Path does not matter; only the change in height of the centre of mass counts.
  • Slow (constant-velocity) lifting or lowering: the agent's work is −Wg-W_g. Lowering a load by hh, the person does −mgh-mgh and gravity does +mgh+mgh.
  • Quick method. Liquid levelling between vessels: only the layer that moves changes height. Work by gravity = its mass × gg × the drop of its centre of mass; ignore the liquid that stays put.
Work by friction and normal force
4× asked, last 2026
Wf=−μkNs,N=mg−Fsin⁡θW_f=-\mu_k N s,\qquad N=mg-F\sin\theta
Kinetic friction, fixed surface (normal does no work); ss = path length. NN: level ground, pull at θ\theta above horizontal.
  • Uniform velocity means zero net work. On a level surface the engine or pull then does work equal to that against friction, μNs\mu Ns (μmgs\mu mgs with no tilted pull); up an incline add mghmgh (hh = rise).
  • A moving surface's normal force does work: the floor of a lift accelerating up at aa does m(g+a)hm(g+a)h on a passenger as it rises by hh.
  • Not μmg\mu mg for a tilted pull. Pulling up at θ\theta cuts NN to mg−Fsin⁡θmg-F\sin\theta (pushing down raises it to mg+Fsin⁡θmg+F\sin\theta). At uniform velocity, solve Fcos⁡θ=μNF\cos\theta=\mu N before finding the work.
  • Not always negative. Static friction on a block carried by a moving incline or truck can do positive work. Use the body's displacement relative to the ground and the angle between it and the friction.
Variable force in one dimension
12× asked, last 2025
W=∫x1x2F(x) dx=area under the F–x graphW=\int_{x_1}^{x_2}F(x)\,dx=\text{area under the } F\text{–}x \text{ graph}
Area below the xx-axis counts negative. Use the given limits, not 0 to xx.
xx (m)FF (N)O2310−5-5+10+10 J−2.5-2.5 J
F=10−5xF=10-5x from 0 to 3 m: work =10−2.5=7.5=10-2.5=7.5 J (area below the axis counts negative)
  • Force of fixed size whose angle to the motion changes with xx: W=∫Fcos⁡θ(x) dxW=\int F\cos\theta(x)\,dx. With θ=kx\theta=kx, from rest at x=0x=0 on a smooth floor, K=(F/k)sin⁡θK=(F/k)\sin\theta.
  • Not k(x2−x1)2/2k(x_2-x_1)^2/2 for a force −kx-kx acting from x1x_1 to x2x_2. xx is measured from the origin, not from where the rough patch starts, so W=−k(x22−x12)/2W=-k(x_2^2-x_1^2)/2.
  • Quick method. Force linear in xx: work = average of the two end forces × distance (a trapezium area). No integration needed.
Variable force in two or three dimensions
3× asked, last 2025
W=∫(Fx dx+Fy dy+Fz dz)W=\int\left(F_x\,dx+F_y\,dy+F_z\,dz\right)
Line integral along the actual path. If FxF_x depends only on xx and FyF_y only on yy, the path does not matter.
  • Quick method. Test: in 2D a force is conservative if ∂Fx/∂y=∂Fy/∂x\partial F_x/\partial y=\partial F_y/\partial x everywhere. Then use W=−ΔUW=-\Delta U or a path parallel to the axes. Integrate each part alone only if FxF_x has no yy and FyF_y no xx.
  • Quick method. If the test fails (e.g. x2y i^x^2y\,\hat i: ∂Fx/∂y=x2\partial F_x/\partial y=x^2, ∂Fy/∂x=0\partial F_y/\partial x=0), the work depends on the path: put the path into each component (y(x)y(x) in FxF_x, x(y)x(y) in FyF_y) before integrating.

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21 more concepts in 5 topics

Work–energy theorem, Conservative forces and potential energy, Spring potential energy, Conservation of mechanical energy and Power. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Work–energy theorem

  • Work–energy theorem14×
  • Work from v(x) or x(t)6×
  • Work by non-conservative forces6×
  • Kinetic energy and momentum9×

Conservative forces and potential energy

  • Conservative force4×
  • Force from potential energy2×
  • Equilibrium and stability3×
  • Reading a U–x graph1×

Spring potential energy

  • Spring force and energy6×
  • Cutting and combining springs
  • Spring compressed between two blocks
  • Body falling onto a spring2×

Conservation of mechanical energy

  • Conservation of mechanical energy7×
  • Connected bodies over a pulley1×
  • Vertical circle: energy view4×
  • Chain on a table1×

Power

  • Power7×
  • Lifts, pumps and efficiency5×
  • Conveyor belt2×
  • Motion at constant power7×
  • Units of energy and power2×

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