SME
University Physics I: Mechanics · Problem type
Energy down a rough slope: heat and the end speed
Use this for “A 40 kg sled starts down a 50 m slope at 20° moving at 3 m/s, with μₖ = 0.08. How much heat does friction make, and how fast is the sled at the bottom?”
Diagram
Weight = m × g = 5 × 9.81 = 49.1 N
Down the slope: Weight sin θ = 49.1 × sin 30° = 24.5 N
Normal force = Weight cos θ = 49.1 × cos 30° = 42.5 N
f = μₖ × normal force = 0.1 × 42.5 = 4.25 N
Net force = Weight sin θ − f = 24.5 − 4.25 = 20.3 N down the slope
a = Net force/m = 20.3/5 = 4.06 m/s²
This is an example. Type your own values to start a new problem.
Q (J)
=
μₖ
×
m (kg)
×
9.81
×
cos(
θ
°)
×
d (m)
The heat Q is the friction force f the picture draws, μₖ times the normal force, times the slope length d.
v₀
Start speed
example
h
Height drop
worked out (not typed)
f
Friction force
worked out (not typed)
v
End speed
worked out (not typed)
Formulas
h
=
d
×
sin(
θ
)
10 = 20 × sin(30°)
f
=
μₖ
×
m
×
9.81
×
cos(
θ
)
4.2479 = 0.1 × 5 × 9.81 × cos(30°)
Q
=
f
×
d
84.9571 = 4.2479 × 20
1
2
×
m
×
v
2
=
1
2
×
m
×
v₀
2
+
m
×
9.81
×
h
−
Q
1
2
×
5
×
12.8925
2
=
1
2
×
5
×
2
2
+
5
×
9.81
×
10
−
84.9571
Assumptions
•
The slope is straight, and the sled slides the whole length d without leaving it; g = 9.81 m/s².
•
Kinetic friction μₖN acts all the way down, where the normal force is N = mg cos θ.
•
Friction’s work turns into heat Q = fd; no other force does work (no push, no air drag).
•
The mass cancels from the end speed: a heavier sled makes more heat but ends just as fast.
Step-by-step
WE KNOW
m = 5 kg, d = 20 m, θ = 30°, μₖ = 0.1, v₀ = 2 m/s
FIND
height drop (h), friction force (f), heat from friction (Q), end speed (v)
Step 1 · Find height drop (h)
Use h = d × sin(θ)
The
slope
is
the
long
side
of
a
right
triangle;
the
drop
is
the
side
opposite
θ
.
h
=
20
×
sin(30°)
h
=
20
×
0.5
h
=
10
m
Step 2 · Find friction force (f)
Use f = μₖ × m × 9.81 × cos(θ)
Into
the
slope
the
forces
balance,
so
N
=
mg
cos
θ
.
Kinetic
friction
is
μₖ
times
N.
f
=
0.1
×
5
×
9.81
×
cos(30°)
f
=
0.1
×
5
×
9.81
×
0.866
f
=
0.5
×
9.81
×
0.866
f
=
4.905
×
0.866
f
=
4.247855
N
Step 3 · Find heat from friction (Q)
Use Q = f × d
Friction
pushes
against
the
motion
the
whole
length
d
,
so
its
work,
−fd,
becomes
heat
fd.
Q
=
4.247855
×
20
Q
=
84.9571
J
Step 4 · Find end speed (v)
Use ½ × m × v² = ½ × m × v₀² + m × 9.81 × h − Q
The
kinetic
energy
at
the
bottom
is
the
start’s,
plus
mgh
from
the
drop,
less
the
heat.
Solve
1
2
mv²
for
v
.
v
=
√(
v₀
2
+
2
×
9.81
×
h
−
2
×
Q
m
)
v
=
√(
2
2
+
2
×
9.81
×
10
−
2
×
84.9571
5
)
v
=
√(4
+
2
×
9.81
×
10
−
2
×
84.9571
5
)
v
=
√(4
+
19.62
×
10
−
169.9142
5
)
v
=
√
4 + 196.2 − 33.98
v
=
√
166.22
v
=
12.8925
m/s
Check
Put the numbers back in. Both sides should match.
10 = 20 × sin(30°) ✓
4.247855 = 0.1 × 5 × 9.81 × cos(30°) ✓
84.9571 = 4.247855 × 20 ✓
1
2
×
5
×
12.8925
2
=
1
2
×
5
×
2
2
+
5
×
9.81
×
10
−
84.9571
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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