SME
Grade 11 Science · Problem type
A block sliding down an incline
Use this for “A 5 kg block slides down a 30° ramp with μₖ = 0.2. What is its acceleration?”
Diagram
Weight = m × g = 5 × 9.8 = 49 N
Down the slope: Weight sin θ = 49 × sin 30° = 24.5 N
Normal force = Weight cos θ = 49 × cos 30° = 42.4 N
f = μₖ × normal force = 0.2 × 42.4 = 8.49 N
Net force = Weight sin θ − f = 24.5 − 8.49 = 16 N down the slope
a = Net force/m = 16/5 = 3.2 m/s²
This is an example. Type your own values to start a new problem.
m
Mass
example
θ
Angle of the ramp
example · °
μₖ
Coefficient of kinetic friction
example
W
Weight
calculated
W∥
Weight along the slope
calculated
FN
Normal force
calculated
f
Friction
calculated
Fₙₑₜ
Net force
calculated
a
Acceleration
calculated · m/s²
Formulas
W
=
m
×
9.8
49 = 5 × 9.8
W
∥
=
W
×
sin(
θ
)
24.5 = 49 × sin(30°)
FN
=
W
×
cos(
θ
)
42.44 = 49 × cos(30°)
f
=
μₖ
×
FN
8.487 = 0.2 × 42.44
Fₙₑₜ
=
W
∥
−
f
16.01 = 24.5 − 8.487
a
=
Fₙₑₜ/
m
3.2 = 16.01/5
Assumptions
•
Tilt the axes with the slope: the weight splits into W sin θ down it and W cos θ into it.
•
Into the slope nothing moves, so FN = W cos θ.
•
The block is sliding down, so kinetic friction acts up the slope; if friction wins, a < 0 and it slows.
Step-by-step
WE KNOW
m = 5 kg, θ = 30°, μₖ = 0.2
FIND
weight (W), weight along the slope (W∥), normal force (FN), friction (f), net force (Fₙₑₜ), acceleration (a)
Step 1 · Find weight (W)
Use W = m × 9.8
Earth pulls each kilogram with 9.8 N.
W
=
5
×
9.8
W
=
49
N
Step 2 · Find weight along the slope (W∥)
Use W∥ = W × sin(θ)
The part of the weight along the slope.
W
∥
=
49
×
sin(30°)
W
∥
=
49
×
0.5
W
∥
=
24.5
N
Step 3 · Find normal force (FN)
Use FN = W × cos(θ)
Into the slope the forces balance: the normal force is the weight’s part into it.
FN = 49 × cos(30°)
FN = 49 × 0.866
FN = 42.44 N
Step 4 · Find friction (f)
Use f = μₖ × FN
Friction is μₖ times the normal force pressing the surfaces together.
f
=
0.2
×
42.435245
f
=
8.487
N
Step 5 · Find net force (Fₙₑₜ)
Use Fₙₑₜ = W∥ − f
Down the slope: the weight’s part less friction.
Fₙₑₜ = 24.5 − 8.487
Fₙₑₜ = 16.01 N
Step 6 · Find acceleration (a)
Use a = Fₙₑₜ/m
Newton’s second law: the net force over the mass.
a
=
16.01/5
a
=
3.2
m/
s
2
Check
Put the numbers back in. Both sides should match.
49 = 5 × 9.8 ✓
24.5 = 49 × sin(30°) ✓
42.44 ≈ 49 × cos(30°) ✓
8.487 ≈ 0.2 × 42.435245 ✓
16.01 ≈ 24.5 − 8.487 ✓
3.2 ≈ 16.01/5 ✓
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