SME
Dynamics
Work–energy and impulse–momentum
Use this for “A 2 kg block moving at 6 m/s, 0.5 m up a smooth ramp, slides down, crosses 2 m of rough floor (μₖ = 0.25) and runs into a spring with k = 800 N/m. How far does it press the spring?”
Diagram
Potential at the start: mgh = 2 × 9.81 × 0.5 = 9.81 J
Kinetic at the start: K₀ = 36 J
Heat on the rough patch: fd = 4.91 × 2 = 9.81 J
Spring: ½kx² = ½ × 800 × 0.3² = 36 J
Energy: ½kx² = K₀ + mgh − fd = 36 + 9.81 − 9.81 = 36 J
This is an example. Type your own values to start a new problem.
1
2
×
m (kg)
×
(
v (m/s)
)
2
+
m (kg)
×
g
×
h (m)
=
μₖ
×
m (kg)
×
g
×
d (m)
+
1
2
×
k (N/m)
×
(
x (m)
)
2
K
Kinetic energy at the start
worked out (not typed) · J
U
Potential energy at the start
worked out (not typed) · J
Wf
Friction work
worked out (not typed) · J
Eₛ
Spring energy
worked out (not typed) · J
Formulas
K
=
1
2
×
m
×
v
2
36
=
1
2
×
2
×
6
2
U
=
m
×
9.81
×
h
9.81 = 2 × 9.81 × 0.5
Wf
=
μₖ
×
m
×
9.81
×
d
9.81 = 0.25 × 2 × 9.81 × 2
Eₛ
=
K
+
U
−
Wf
36 = 36 + 9.81 − 9.81
Eₛ
=
1
2
×
k
×
x
2
36
=
1
2
×
800
×
0.3
2
Assumptions
•
Friction acts only on the rough level patch, and its work Wf = μkmgd turns to heat.
•
The spring is light and ideal: pressed a distance x, it stores ½kx², and the block stops there.
•
Work–energy: K + mgh = Wf + ½kx², with K and h at the start and g = 9.81 m/s².
Step-by-step
WE KNOW
m = 2 kg, v = 6 m/s, h = 0.5 m, μₖ = 0.25, d = 2 m, k = 800 N/m
FIND
kinetic energy at the start (K), potential energy at the start (U), friction work (Wf), spring energy (Eₛ), compression (x)
Step 1 · Find kinetic energy at the start (K)
Use K = ½ × m × v²
The kinetic energy the block starts with.
K
=
1
2
×
2
×
6
2
K
=
1
2
×
2
×
36
K
=
36
J
Step 2 · Find potential energy at the start (U)
Use U = m × 9.81 × h
The potential energy at the start, measured from the floor.
U
=
2
×
9.81
×
0.5
U
=
19.62
×
0.5
U
=
9.81
J
Step 3 · Find friction work (Wf)
Use Wf = μₖ × m × 9.81 × d
Friction
on
level
ground
is
μₖ
times
the
weight
mg;
times
the
length
d
it
acts
over,
it
is
the
work
turned
to
heat.
Wf = 0.25 × 2 × 9.81 × 2
Wf = 0.5 × 9.81 × 2
Wf = 4.905 × 2
Wf = 9.81 J
Step 4 · Find spring energy (Eₛ)
Use Eₛ = K + U − Wf
Work–energy: what the block had, less the heat on the patch, is stored in the spring.
Eₛ = 36 + 9.81 − 9.81
Eₛ = 36 J
Step 5 · Find compression (x)
Use Eₛ = ½ × k × x²
Double
the
energy,
divide
by
k
,
and
take
the
square
root;
a
compression
is
positive.
x
=
√(2
×
Eₛ
k
)
x
=
√(2
×
36
800
)
x
=
√(
72
800
)
x
=
√
0.09
x
=
0.3
m
Check
Put the numbers back in. Both sides should match.
36
=
1
2
×
2
×
6
2
✓
9.81 = 2 × 9.81 × 0.5 ✓
9.81 = 0.25 × 2 × 9.81 × 2 ✓
36 = 36 + 9.81 − 9.81 ✓
36
=
1
2
×
800
×
0.3
2
✓
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