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A box sliding down a rough ramp

The problem

A 5.0 kg box is released from rest at the top of a ramp tilted 30° above horizontal. The coefficient of kinetic friction between the box and the ramp is 0.20. The box slides 4.0 m down the ramp.

The question

  1. aWhat is the box’s acceleration down the ramp?find a
  2. bHow fast is it moving after sliding 4.0 m?find v
1

Set up the problem

Sort everything into two lists: what you know, and what you need to find.

Givens

  • m = 5.0 kg
  • θ = 30°
  • μk = 0.20
  • d = 4.0 m
  • v0 = 0hidden in “released from rest”
  • g = 9.8 m/s²not in the problem, but always there

Find

  • Part (a): find a the acceleration down the ramp
  • Part (b): find v the speed after sliding 4.0 m

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2

Draw the forces and tilt your axes

Only three forces act on the box. Weight mg pulls straight down. The normal force N pushes out of the ramp, perpendicular to its surface. Kinetic friction f acts along the surface, opposite the motion, so it points up the ramp.

Put +x along the ramp, pointing downhill, and +y perpendicular to the ramp.

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3

Split the weight into two parts

Along the ramp: mg sin θ = (5.0)(9.8)(sin 30°) = 24.5 N
Into the ramp: mg cos θ = (5.0)(9.8)(cos 30°) = 42.4 N

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4

Calculate the normal force

Nothing accelerates perpendicular to the ramp, so the y forces balance:

ΣFy = N − mg cos θ = 0
N = mg cos θ = 42.4 N

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5

Calculate the friction force

f = μkN = (0.20)(42.4) = 8.49 N, up the ramp

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6

Find the acceleration

Apply Newton’s second law along the ramp:

ΣFx = mg sin θ − f = ma
a = g(sin θ − μk cos θ)
a = 9.8(0.500 − 0.173) = 3.20 m/s² down the ramp

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7

Find the final speed

The acceleration is constant, so kinematics applies. No time is given and none is asked for:

v² = v0² + 2ad = 0 + 2(3.20)(4.0) = 25.6
v = 5.06 m/s

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Final answer

(a) a = 3.20 m/s² down the ramp

(b) v = 5.06 m/s after 4.0 m

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Try a similar problem on your own

The problem

A 2.0 kg sled starts from rest on a snowy hill inclined at 25°. The coefficient of kinetic friction is 0.15. The sled slides 6.0 m down the hill.

The question

  1. aWhat is the sled’s acceleration?find a
  2. bWhat is its speed after 6.0 m?find v

Follow the same seven steps. Then check yourself:

m/s²

m/s

Stuck? Get a hint for each step

Normal force: it’s not mg. Use the part of the weight that pushes into the hill.

Friction: multiply μk by the normal force you just found.

Acceleration: a = g(sin θ − μk cos θ). Does the mass matter?

Speed: no time given, so use the equation without t. Remember “starts from rest.”

Show the answers

N = 17.8 N, f = 2.66 N, a = 2.81 m/s², v = 5.81 m/s

Your turn

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Put it into practice with the simulation

Change the ramp and watch what happens. Before you press release, predict the acceleration, then check it.

a = 0 m/s² t = 0.00 s slid 0.0 m v = 0.00 m/s

Keep going

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