Notes
Forces
- A force is a push or pull measured in newtons (N). Types: weight, normal, tension, friction, applied, air resistance.
- A free-body diagram shows every force on one object. The net force is their vector sum.
Newton's three laws
- First law: with zero net force, velocity stays constant (including at rest). Inertia depends on mass.
- Second law: $F_{\text{net}} = ma$.
- Third law: forces come in equal and opposite pairs acting on different objects.
Weight, normal force and friction
- Weight $W = mg$; mass stays the same everywhere.
- On a flat surface, $N = mg$. In an accelerating elevator, $N = m(g \pm a)$.
- Friction: $f_k = \mu_k N$; static friction can be up to $\mu_s N$.
Inclines
- Along the ramp: $mg\sin\theta$. Perpendicular: $N = mg\cos\theta$. Frictionless: $a = g\sin\theta$.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$F = ma = 1500 \times 2 = 3000 \text{ N}$.
Constant velocity means zero net force, so the forces balance.
$\displaystyle m = \frac{W}{g} = \frac{49}{9.8} = 5 \text{ kg}$.
By Newton's third law, the gas exerts an equal and opposite force on the rocket.
Friction $= 0.25 \times 30 \times 9.8 = 73.5 \text{ N}$; net $= 26.5 \text{ N}$; $\displaystyle a = \frac{26.5}{30} \approx 0.88 \text{ m/s}^2$.
$N = mg\cos 30^\circ = 196 \times 0.866 \approx 170 \text{ N}$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A worker pulls a 40 kg crate across a level warehouse floor with a horizontal rope. The tension in the rope is 200 N and the coefficient of kinetic friction is 0.35.
(a) List the forces on the crate and find the normal force.
Show the answer and scoring guide
Forces: weight (down), normal force (up), tension (forward), kinetic friction (backward). Vertical forces balance: $N = mg = 40 \times 9.8 = 392 \text{ N}$.
Check off the points you earned:
(b) Find the friction force and the crate's acceleration.
Show the answer and scoring guide
$f_k = 0.35 \times 392 = 137.2 \text{ N}$. $F_{\text{net}} = 200 - 137.2 = 62.8 \text{ N}$. $\displaystyle a = \frac{62.8}{40} \approx 1.57 \text{ m/s}^2$.
Check off the points you earned:
(c) What tension would make the crate move at constant velocity? Explain using Newton's first law.
Show the answer and scoring guide
At constant velocity the net force is zero, so tension must equal friction: $T = 137.2 \text{ N}$.
Check off the points you earned: