Physics Sensei practice › forces

Practice: a box sliding down a rough ramp (Algebra-Based)

Three problems, each a little different from the last. Answer first and check. Open the steps only when you get stuck.

Want to see one solved step by step first? A box sliding down a rough ramp

1

Same problem, new numbers

The problem

An 8.0 kg crate is released from rest at the top of a ramp tilted 35° above horizontal. The coefficient of kinetic friction is 0.25. The crate slides 5.0 m down the ramp.

The question

  1. aWhat is the crate’s acceleration down the ramp?find a
  2. bHow fast is it moving after sliding 5.0 m?find v

Try it first. Then check.

m/s²

m/s

Show the steps

What are they asking?

(a) an acceleration, so forces and Newton’s second law. (b) a speed after a distance, with no time given, so v² = v₀² + 2ad.

What do I know?

m = 8.0 kg, θ = 35°, μk = 0.25, d = 5.0 m, v₀ = 0 (hidden in “released from rest”), g = 9.8 m/s² (always there).

What’s the big idea?

Newton’s second law along the ramp: the weight’s part down the ramp minus friction equals ma.

Do the math

N = mg cos θ = (8.0)(9.8)(cos 35°) = 64.2 N
f = μkN = (0.25)(64.2) = 16.1 N
a = (mg sin θ − f)/m = (45.0 − 16.1)/8.0 = 3.61 m/s²
v = √(2ad) = √(2 · 3.61 · 5.0) = 6.01 m/s

Does it make sense?

Units work out (m/s² and m/s). The acceleration is positive and smaller than g sin θ, as it should be with friction.

2

Same idea, one twist

The problem

A 3.0 kg box is shoved up a ramp tilted 30° with a starting speed of 6.0 m/s. The coefficient of kinetic friction is 0.20. The box slides up, slows down and stops.

The question

  1. aHow fast is the box slowing down while it slides up? (Give the size of the acceleration.)find a
  2. bHow far up the ramp does it go before it stops?find d

Try it first. Then check.

m/s²

m

Show the steps

What are they asking?

(a) how fast it slows down: forces again. (b) a distance until it stops, with no time given: v² = v₀² + 2ad.

What do I know?

m = 3.0 kg, θ = 30°, μk = 0.20, v₀ = 6.0 m/s, v = 0 at the end (hidden in “stops”). The box moves up, so friction points down the ramp.

What’s the big idea?

Newton’s second law along the ramp, with both forces pointing down the ramp this time.

Do the math

mg sin θ = (3.0)(9.8)(sin 30°) = 14.7 N
f = μkmg cos θ = (0.20)(3.0)(9.8)(cos 30°) = 5.09 N
a = (14.7 + 5.09)/3.0 = 6.60 m/s² (slowing down)
d = v₀²/(2a) = (6.0)²/(2 · 6.60) = 2.73 m

Does it make sense?

Units work out (m). Plus sign for friction, not minus: if your slowdown came out smaller than 4.90 m/s² (gravity alone), friction was pointing the wrong way.

3

Spot the idea

The problem

A book rests on a board. You slowly tilt the board. The coefficient of static friction between the book and the board is 0.50.

The question

  1. aAt what angle does the book start to slide?find θ

Try it first. Then check.

degrees

Show the steps

What are they asking?

An angle, at the moment the book is about to slide. Nothing is accelerating yet.

What do I know?

μs = 0.50. “About to slide” means static friction is at its maximum: f = μsN. The forces still balance, so a = 0.

What’s the big idea?

Balanced forces (Newton’s first law), not F = ma with an acceleration. Same ramp, different idea.

Do the math

mg sin θ = μsmg cos θ
tan θ = μs = 0.50
θ = tan⁻¹(0.50) = 26.6°

Does it make sense?

Between 0° and 90°, and the mass canceled, which matches what you see: a heavy book and a light book start sliding at the same angle.