NLM Graphs
NCERT · Class XI · Physics

Newton's Laws Graphs

A purely visual and mathematical breakdown of all possible graphs in Newton's Laws of Motion. No general theory—just the exact coordinate systems, slopes, and areas required to solve complex mechanics problems. Designed explicitly for targeted high-yield visual revision.

7
Essential Graphs
100%
Visual Focus
High
Exam Yield
01

Force vs Time (F-t) Graph

Visualizing impulsive forces and total change in momentum over a time interval.

t F Fmax Area = Δp
Force vs Time Graph
The area enclosed under the F-t curve equals the Impulse ($J$), which is exactly the total Change in Momentum ($\Delta p$) delivered to the body.
02

Momentum vs Time (p-t) Graph

Deriving instantaneous applied force from the rate of change of momentum.

t p dt dp Slope = Fnet
Momentum vs Time Graph
The slope of the tangent at any point on the p-t curve ($\frac{dp}{dt}$) gives the instantaneous Net Force acting on the body at that moment.
03

Force vs Position (F-x) Graph

Calculating work done by conservative forces like an ideal spring.

x F +W −W F = −kx
Force vs Position (F-x) Graph
A straight line through the origin with negative slope represents a restoring Spring Force ($F = -kx$). The area between the line and the x-axis gives the Work Done, positive on one side and negative on the other.
04

Friction vs Applied Force (f-F) Graph

The most vital graph for understanding static, limiting, and kinetic friction phases.

F f 45° fs,max fk Static: f = F f = fk (const)
Friction vs Applied Force Graph
Friction matches applied force exactly in the static region (a 45° line, since $f = F$) up to the limiting value $f_{s,max}$, then drops slightly and stays constant at the kinetic value $f_k$.
05

Acceleration vs Applied Force (a-F) Graph

Mapping how an object accelerates on a rough surface once friction is overcome.

F a fk Slope = 1/m a = 0 (F ≤ f_k)
Acceleration vs Applied Force (a-F)
An object on a rough surface doesn't accelerate until the applied Force overcomes friction. The x-intercept marks the kinetic friction $f_k$, and the slope of the rising line equals $1/m$.
06

Apparent Weight vs Acceleration (Lift Problem)

The single most-asked NLM application graph — a person standing on a weighing machine inside a lift.

mg N = 0 −g N = m(g + a) a N
Apparent Weight vs Acceleration
Taking upward acceleration as positive, the Normal reaction is $N = m(g + a)$ — a straight line. At $a = 0$ the reading is the true weight $mg$; at $a = -g$ (a snapped cable, free fall) the reading drops to zero, giving true weightlessness.
07

Acceleration vs 1/Mass (a vs 1/m)

Newton's Second Law re-plotted to isolate mass as a straight-line relationship.

1/m a Slope = F₁ Slope = F₂ F₂ > F₁
Acceleration vs 1/Mass Graph
Since $a = F \cdot \frac{1}{m}$ for constant force, plotting $a$ against $1/m$ gives a straight line through the origin. A steeper line means a larger constant force was applied.

Quick Revision Sheet

Formulas explicitly tied to graphing mechanics problems.

Area under F-t Graph
$$ J = \int F dt = \Delta p $$
Slope of p-t Graph
$$ \frac{dp}{dt} = F_{net} $$
Area under F-x Graph
$$ W = \int F dx = \Delta K $$
Slope of a-F Graph
$$ m = \frac{1}{\text{Slope}} $$
x-intercept of a-F Graph
Kinetic Friction ($f_k$)
Static Friction Region Slope
$$ m = 1 \ (\theta = 45^\circ) $$
Apparent Weight in Lift
$$ N = m(g \pm a) $$
Free Fall / Weightlessness
At $a = -g$, $N = 0$
Slope of a vs 1/m Graph
$$ \text{Slope} = F \ (\text{constant}) $$