SME
Grade 11 Science · Physics
Work, energy, power and simple machines
Use this for “A 0.25 kg ball drops from 8 m. How fast is it going at 3 m?” with conservation of energy.
Diagram
U + K = 7.35 J + 12.25 J = 19.6 J
U = m × g × h
U
=
0.25
kg
×
9.8
m/
s
2
×
3
m
U = 7.35 J
K
=
1
2
×
m
×
v
2
K
=
1
2
×
0.25
kg
×
(9.9 m/s)
2
K = 12.25 J
Going down, potential energy turns into kinetic energy; the total stays the same
This is an example. Type your own values to start a new problem.
m (kg)
×
9.8
m/s²
×
h₀ (m)
=
m (kg)
×
9.8
m/s²
×
h (m)
+
½
×
m (kg)
×
(
v (m/s)
)
2
U
Potential energy
calculated · J
K
Kinetic energy
calculated · J
E
Total energy
calculated · J
Formulas
E
=
m
×
9.8
×
h₀
19.6 = 0.25 × 9.8 × 8
U
=
m
×
9.8
×
h
7.35 = 0.25 × 9.8 × 3
K
=
E
−
U
12.25 = 19.6 − 7.35
K
=
1
2
×
m
×
v
2
12.25
=
1
2
×
0.25
×
9.9
2
Assumptions
•
The car starts at rest at height h₀, and heights are measured from the lowest point.
•
No friction or air resistance, so the total energy E stays the same: U + K = E.
•
Mass cancels out of the speed: every car from the same height has the same v there.
Step-by-step
WE KNOW
m = 0.25 kg, h₀ = 8 m, h = 3 m
FIND
total energy (E), potential energy (U), kinetic energy (K), speed now (v)
Step 1 · Find total energy (E)
Use E = m × 9.8 × h₀
At the start it is all potential energy.
E
=
0.25
×
9.8
×
8
E
=
2.45
×
8
E
=
19.6
J
Step 2 · Find potential energy (U)
Use U = m × 9.8 × h
Each kilogram lifted each meter stores 9.8 J.
U
=
0.25
×
9.8
×
3
U
=
2.45
×
3
U
=
7.35
J
Step 3 · Find kinetic energy (K)
Use K = E − U
What is not potential energy is kinetic energy.
K
=
19.6
−
7.35
K
=
12.25
J
Step 4 · Find speed now (v)
Use K = ½ × m × v²
Double the kinetic energy, divide by the mass, take the square root.
v
=
√
2 ×
K
m
v
=
√
2 × 12.25/0.25
v
=
√
24.5/0.25
v
=
√
98
v
=
9.9
m/s
Check
Put the numbers back in. Both sides should match.
19.6 = 0.25 × 9.8 × 8 ✓
7.35 = 0.25 × 9.8 × 3 ✓
12.25 = 19.6 − 7.35 ✓
12.25
≈
1
2
×
0.25
×
9.9
2
✓
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Work done by a pull at an angle
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