SME
Grade 11 Science · Problem type
A spring launcher, friction and a ramp
Use this for “A spring (k = 400 N/m) pressed 0.15 m launches a 0.5 kg block over 0.5 m of floor with 2 N of friction. How high up the smooth ramp does it go?”
Diagram
Spring: ½kx² = ½ × 400 × 0.15² = 4.5 J
Heat on the rough patch: fd = 2 × 0.5 = 1 J
Potential: mgh = 0.5 × 9.8 × 0.714 = 3.5 J
Kinetic: 4.5 − 1 − 3.5 = 0 J
This is an example. Type your own values to start a new problem.
½
×
k (N/m)
×
(
x (m)
)
2
=
f (N)
×
d (m)
+
m (kg)
×
9.8
m/s²
×
h (m)
+
K (J)
Eₛ
Spring energy
calculated · J
Q
Heat
calculated · J
U
Potential energy
calculated · J
Formulas
Eₛ
=
1
2
×
k
×
x
2
4.5
=
1
2
×
400
×
0.15
2
Q
=
f
×
d
1 = 2 × 0.5
U
=
m
×
9.8
×
h
3.5 = 0.5 × 9.8 × 0.714
K
=
Eₛ
−
Q
−
U
0 = 4.5 − 1 − 3.5
Assumptions
•
The spring stores ½kx²; friction turns fd into heat; the smooth ramp trades kinetic energy for mgh.
•
The total never changes: spring energy = heat + potential + kinetic.
•
At the highest point the block stops for an instant: type K = 0 to find that height.
Step-by-step
WE KNOW
k = 400 N/m, x = 0.15 m, m = 0.5 kg, f = 2 N, d = 0.5 m, K = 0 J
FIND
spring energy (Eₛ), heat (Q), potential energy (U), height on the ramp (h)
Step 1 · Find spring energy (Eₛ)
Use Eₛ = ½ × k × x²
A
spring
pressed
x
stores
1
2
kx².
Eₛ
=
1
2
×
400
×
0.15
2
Eₛ
=
1
2
×
400
×
0.0225
Eₛ = 200 × 0.0225
Eₛ = 4.5 J
Step 2 · Find heat (Q)
Use Q = f × d
Friction’s work on the rough patch becomes heat: force times distance.
Q
=
2
×
0.5
Q
=
1
J
Step 3 · Find potential energy (U)
Use K = Eₛ − Q − U
What is left after the heat and the kinetic energy.
U
=
Eₛ
−
Q
−
K
U
=
4.5
−
1
−
0
U
=
3.5
J
Step 4 · Find height on the ramp (h)
Use U = m × 9.8 × h
Divide the potential energy by mg.
h
=
U
/(
m
×
9.8)
h
=
3.5/(0.5
×
9.8)
h
=
3.5/4.9
h
=
0.714
m
Check
Put the numbers back in. Both sides should match.
4.5
=
1
2
×
400
×
0.15
2
✓
1 = 2 × 0.5 ✓
3.5 ≈ 0.5 × 9.8 × 0.714 ✓
0 = 4.5 − 1 − 3.5 ✓
Topic check
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