Work, Energy and Power formulas

Class 11 physics formula sheet for NEET and JEE: the key equations of NCERT chapter 5, the special cases questions are built on, and diagrams where they help.

23 formulas6 sectionsClass 11 · Chapter 52 of 6 sections free

By Sreeraj P, M.Sc Physics · 10+ years teaching NEET and JEE

Most used formulasOther formulas and cases

1Work

$$\begin{array}{l}\displaystyle W=\vec F\cdot\vec s=Fs\cos\theta\\[5pt]\displaystyle W=\int\vec F\cdot d\vec s=\int F_xdx+\int F_ydy+\int F_zdz\end{array}$$

Area under $F$–$x$ graph (below axis negative). Force varying linearly $F_1\to F_2$: $\frac{F_1+F_2}2s$. Zero work: $F\perp s$ (centripetal force, tension in a pendulum, porter on level road) or $s=0$. 1 J $=10^7$ erg; 1 kWh $=3.6\times10^6$ J.

Case	Work
Rope lifting $m$ with acceleration $a$ up / lowering	$m(g+a)h$ / $-m(g-a)h$
Pendulum pulled to angle $\theta$ by horizontal force	$W_g=-mgL(1-\cos\theta)$, $W_F=FL\sin\theta$ (if $F$ constant)
Rod turned through $\theta$ about one end	$W_g=-mg\frac l2(1-\cos\theta)$; ladder raised to $\theta$: $mg\frac L2\sin\theta$
Hanging $1/n$ of chain pulled onto table	$\dfrac{MgL}{2n^2}$; from $1/n_1$ to $1/n_2$: $\dfrac{MgL}2\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right)$
Lower end of hanging chain lifted to top	$\dfrac{MgL}4$
Bucket $M$ with rope $m$ from depth $l$	$Mgl+\dfrac{mgl}2$
Lifting body (density $d_1$) in liquid ($d_2$)	$mgh\left(1-\dfrac{d_2}{d_1}\right)$
Cylinder turned from side to end	$mg\left(\dfrac l2-r\right)$
Stacking $n$ bricks (height $h$)	$\dfrac{n(n-1)}2mgh$
Gas	$W=\int P\,dV$
Tension on Atwood masses (together)	0 (equal and opposite)

2Energy

$$K=\tfrac12mv^2=\frac{p^2}{2m},\qquad p=\sqrt{2mK}$$

Same $p$: $K\propto1/m$ (bullet has more KE than gun). Same $K$: $p\propto\sqrt m$. Momentum up $x\%$ (small): KE up $2x\%$; momentum doubled → KE ×4; KE ×$n$ → $p\times\sqrt n$. $K$ vs $p$: parabola; $\sqrt K$ vs $p$: straight line.

$$\begin{array}{l}\displaystyle F=-\frac{dU}{dx},\quad \vec F=-\left(\frac{\partial U}{\partial x}\hat i+\frac{\partial U}{\partial y}\hat j+\frac{\partial U}{\partial z}\hat k\right)\\[5pt]\displaystyle \Delta U=-W_{\text{cons}}\end{array}$$

$$\begin{array}{l}\displaystyle F=-\frac{dU}{dx}\\[6pt]\displaystyle \vec F=-\left(\frac{\partial U}{\partial x}\hat i+\frac{\partial U}{\partial y}\hat j+\frac{\partial U}{\partial z}\hat k\right)\\[6pt]\displaystyle \Delta U=-W_{\text{cons}}\end{array}$$

PE only for conservative forces (gravity, spring, electrostatic): work around a closed path is zero and path-independent. Friction, viscosity: non-conservative.

$$U_{\text{spring}}=\tfrac12kx^2,\qquad W_{x_1\to x_2}=\tfrac12k(x_2^2-x_1^2)=\tfrac12k\,y(2x_1+y)$$

$$\begin{array}{l}\displaystyle U_{\text{spring}}=\tfrac12kx^2\\[6pt]\displaystyle W_{x_1\to x_2}=\tfrac12k(x_2^2-x_1^2)=\tfrac12k\,y(2x_1+y)\end{array}$$

Extra stretch $y$ after $x_1$. Spring force $-kx$. Rod standing vertically: $U=\dfrac{mgL}2$. Chain on hemisphere (length $l$, one end at top): $U=\dfrac{mgR^2}l\sin\dfrac lR$.

Equilibrium $\dfrac{dU}{dx}=0$. Stable: $U$ minimum, $\dfrac{d^2U}{dx^2}>0$. Unstable: maximum. Neutral: $U$ constant.
$U=\dfrac a{x^{12}}-\dfrac b{x^6}$: equilibrium at $x=\left(\dfrac{2a}b\right)^{1/6}$, $U_{\min}=-\dfrac{b^2}{4a}$.

4 more sections and 14 formulas in the full chapter

3Work–energy theorem and conservation2 formulas · 1 case table
4Power2 formulas · 1 case table
5Vertical circle5 formulas · 2 case tables · 1 diagram
6Collisions5 formulas · 2 case tables

Get the complete Work, Energy and Power sheet, with the whole Class 11 Physics Formula Book (all chapters, as a PDF), free on WhatsApp.

Get the full PDF on WhatsApp

Practise these formulas54 Work, Energy and Power questions with step-by-step solutions →