SME
Dynamics · Problem type
Power and efficiency of a hoist
Use this for “A hoist raises a 500 kg crate 12 m at a steady 0.8 m/s, and its motor is 85% efficient (η = 0.85). What output power lifts the crate, and what input power does the motor draw?”
Diagram
Work done: W = mgh = 500 × 9.81 × 12 = 58,860 J
Power: Pₒᵤₜ = W/t = 58,860/15 = 3,924 W
A watt is a joule every second: the same work done faster needs more power.
This is an example. Type your own values to start a new problem.
Pₒᵤₜ (W)
=
m (kg)
×
g
×
v (m/s)
Pᵢₙ (W)
=
Pₒᵤₜ (W)
÷
η
h
Height
example
t
Time
calculated
W
Work
calculated
Formulas
v
=
h
t
0.8
=
12
15
W
=
m
×
9.81
×
h
58,860 = 500 × 9.81 × 12
Pₒᵤₜ
=
m
×
9.81
×
v
3,924 = 500 × 9.81 × 0.8
Pᵢₙ
=
Pₒᵤₜ
η
4,616.4706
=
3,924
0.85
Assumptions
•
The crate rises at a steady speed, so the hoist pulls with exactly its weight mg (g = 9.81 m/s²).
•
Output power is force times speed, Pₒᵤₜ = mgv: the work mgh spread over the time t.
•
Efficiency η = Pₒᵤₜ ÷ Pᵢₙ: the motor draws more than it delivers, and the rest is lost as heat.
Step-by-step
WE KNOW
v = 0.8 m/s, m = 500 kg, h = 12 m, η = 0.85
FIND
time (t), work (W), output power (Pₒᵤₜ), input power (Pᵢₙ)
Step 1 · Find time (t)
Use v = h ÷ t
Divide the height by the speed.
t
=
h
v
t
=
12
0.8
t
=
15
s
Step 2 · Find work (W)
Use W = m × 9.81 × h
The pull, equal to the weight mg, times the height it acts through.
W
=
500
×
9.81
×
12
W
=
4,905
×
12
W
=
58,860
J
Step 3 · Find output power (Pₒᵤₜ)
Use Pₒᵤₜ = m × 9.81 × v
Power
is
force
times
velocity:
the
pull
mg
times
the
steady
speed
v
(the
same
as
W
t
).
Pₒᵤₜ = 500 × 9.81 × 0.8
Pₒᵤₜ = 4,905 × 0.8
Pₒᵤₜ = 3,924 W
Step 4 · Find input power (Pᵢₙ)
Use Pᵢₙ = Pₒᵤₜ ÷ η
η
=
Pₒᵤₜ
Pᵢₙ
,
so
the
motor
draws
the
output
power
divided
by
the
efficiency.
Pᵢₙ
=
3,924
0.85
Pᵢₙ = 4,616.471 W
Check
Put the numbers back in. Both sides should match.
0.8
=
12
15
✓
58,860 = 500 × 9.81 × 12 ✓
3,924 = 500 × 9.81 × 0.8 ✓
4,616.471
=
3,924
0.85
✓
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