Physics 1 · Forces test

Motion Test Prep

Inclined planes and circular motion. Short lessons, flashcards, and practice that reveals one step at a time. g = 9.8 m/s² everywhere.

Plan
  1. Read both lessons. Cover each worked example and try it before reading the steps.
  2. Practice tab: every problem on paper first, then reveal steps.
  3. Flashcards until every card is "Got it".
  4. Redo anything you got wrong, then stop and rest before the test.
Lesson 1

Inclined planes

Every incline problem is the same move: tilt your axes so x runs along the slope. Then only gravity needs splitting into parts.

θ +x+y tilted axes θ mg N f mg sin θ mg cos θ
gravity mg, straight downnormal N, out of the surfacefriction f, against the motion

The two directions

Along the slope: mg sin θ pulls it down the slope
Into the slope: mg cos θ, cancelled by N → N = mg cos θ
Friction: fk = μk N = μk mg cos θ sliding
Static: fs ≤ μs N only equals μsN when about to slip

The answers you'll keep getting

No friction: a = g sin θ mass cancels
Sliding down with friction: a = g (sin θ − μk cos θ)
Sliding up (coasting): a = g (sin θ + μk cos θ), slowing it down
About to slip, or sliding at constant speed: μ = tan θ

Once you have a, it's kinematics along the slope: v² = v₀² + 2ad, with d measured along the ramp. Height of the ramp is h = d sin θ.

Traps: N is NOT mg on a slope, it's mg cos θ. Mixed up sin and cos? Imagine θ = 0 (flat ground): the pull along the slope must vanish, so that one is sin θ. Friction always points against the motion, so it flips when the block moves up the slope. A block sitting still has fs = mg sin θ, not μsN.
Example: 5.0 kg block on a frictionless 30° incline. Find a and N.
  1. a = g sin θ = 9.8 × sin 30° = 4.9 m/s² down the slope
  2. N = mg cos θ = 5.0 × 9.8 × cos 30° = 42.4 N
Example: 2.0 kg box slides down a 25° incline, μk = 0.20. Find a.
  1. Along slope: mg sin θ − μk mg cos θ = ma
  2. m cancels: a = 9.8 (sin 25° − 0.20 cos 25°) = 9.8 (0.423 − 0.181)
  3. a = 2.36 m/s² down the slope
Lesson 2

Circular motion

Constant speed in a circle still means accelerating, because the direction keeps changing. That acceleration points to the center, so the net force does too.

direction of travel r v ac v ac same speed, new direction
velocity v, along the circleacceleration ac, to the center
ac = v² / r = ω² r points to the center
v = 2πr / T  ·  ω = 2π / T = v / r  ·  f = 1 / T
ΣF (toward center) = m v² / r

Where the center force comes from

"Centripetal force" is never a new force. Draw the real forces, take toward the center as +, and set the sum equal to mv²/r.

Ball on a string (horizontal)
FT = mv²/r
Car on a flat curve
fs = mv²/r → vmax = √(μs g r)
Banked curve, no friction
N cos θ = mg, N sin θ = mv²/r → tan θ = v²/(rg)
Top of a loop
N + mg = mv²/r → slowest: v = √(g r)
Bottom of a loop or dip
N − mg = mv²/r → N bigger than mg
Top of a hill
mg − N = mv²/r → N smaller than mg
Traps: rpm or rev/s means find T first (2 rev/s → T = 0.5 s). Radius, not diameter. A banked curve is NOT an incline problem: the acceleration is horizontal, so keep axes horizontal and vertical, and N is bigger than mg cos θ. There is no outward "centrifugal" force on your diagram.
Example: car on a flat curve, r = 50 m, μs = 0.80. Max speed?
  1. Friction is the center force: μs mg = mv²/r
  2. m cancels: v = √(μs g r)
  3. v = √(0.80 · 9.8 · 50) = 19.8 m/s
Example: 60 kg rider at the top of a loop, r = 10 m, v = 12 m/s. Force from the seat?
  1. At the top both N and mg point down, toward the center
  2. N + mg = mv²/r → N = m (v²/r − g)
  3. N = 60 (14.4 − 9.8) = 276 N