SME
Grade 11 Science · Problem type
Power: work done each second
Use this for “A 60 kg student runs up 4.5 m of stairs in 6 s. How much work does she do, and what is her power?”
Diagram
Work done: W = mgh = 60 × 9.8 × 4.5 = 2,646 J
Power: P = W/t = 2,646/6 = 441 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.
P (W)
=
m (kg)
×
9.8
m/s²
×
h (m)
t (s)
Fg
Weight
calculated
W
Work done
calculated · J
Formulas
Fg
=
m
×
9.8
588 = 60 × 9.8
W
=
Fg
×
h
2,646 = 588 × 4.5
P
=
W
t
441
=
2,646
6
Assumptions
•
Lifting at a steady speed takes a force equal to the weight, Fg = mg, over the height: W = Fg h = mgh.
•
Power is work per second: P = W/t. One watt is one joule each second.
•
The path does not matter, only the height gained: stairs or a rope, the work is the same.
•
Given a weight in newtons instead of a mass, type it as Fg.
Step-by-step
WE KNOW
m = 60 kg, h = 4.5 m, t = 6 s
FIND
weight (Fg), work done (W), power (P)
Step 1 · Find weight (Fg)
Use Fg = m × 9.8
The weight is the mass times g.
Fg = 60 × 9.8
Fg = 588 N
Step 2 · Find work done (W)
Use W = Fg × h
The weight, mg, times the height raised.
W
=
588
×
4.5
W
=
2,646
J
Step 3 · Find power (P)
Use P = W/t
The joules for each second.
P
=
2,646
6
P
=
441
W
Check
Put the numbers back in. Both sides should match.
588 = 60 × 9.8 ✓
2,646 = 588 × 4.5 ✓
441
=
2,646
6
✓
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