1. Power

2. Key ideas Power is the rate at which work is done
Power = Force x velocity
Power = Fv
and Power = work done ÷ time taken
P =
Power is measured in watts (W)
You will probably use kilowatts 1kW = 1000W

3. How to answer the questions
Draw a clear diagram
Summarise the information given
It may be necessary to calculate the work done using KE = ½mv²,
PE = mgh and W = Fs
Use P = Fv or P =
You may also need to resolve forces or to apply F = ma where F is the resultant force

4. Find the magnitude of the resistance forces at this speed
Example 1
A cyclist travels at a constant speed of 10ms-1 along a straight horizontal road. The power the cyclist develops is 400W and the mass of the cyclist and bicycle totals 80 kg.
Find the magnitude of the resistance forces at this speed
The resistance forces are found to be proportional to the cyclist’s speed
b Write down the equation for the resistance forces
C Calculate the acceleration of the cyclist when his speed is 2 ms-1 if the power developed remains unchanged
Draw a clear diagram then click to continue
normal reaction
10 ms-1 R F
80g

5. Now summarise the information
P = 400 W F = ? v = 10 ms-1 a.
Apply P = Fv
P = Fv
400 = F x 10
F = 40 N
b. R  v
So R = kv k is the constant of proportion
because the speed is constant R = 40N
40 = k x 10
k = 4
R = 4v

6. Apply F = ma remember this F is the resultant force 200 – 8 = 80a
normal reaction
2 ms-1 R F
80g
R = ? V = 2 ms-1
R = 4v
R = 4 x 2
R = 8N
P = Fv
400 = F x 2
F = 200 N
Apply F = ma remember this F is the resultant force
200 – 8 = 80a
80a = 192
a = 2.4 ms-2

7. The work done by the pump The power developed by the pump
Example 2
A pump raises 20 kg of water to a height of 5m in 30 seconds. The water starts at rest and it is issued at 5ms-1. Calculate:
The work done by the pump
The power developed by the pump
State one key assumption you have made.
5 ms-1
20 kg delivered in 30 seconds
0 ms-1
5 m
pump
zero PE level
The work done by the pump is the same as the increase in mechanical energy.
W = KEend + PEend
W = mgh + ½mv²
W = (20 x 9.8 x 5) + (½ x 20 x 5²)
W = 1230 J

8. We assume the force produced by the pump is constant
5 ms-1
20 kg delivered in 30 seconds
0 ms-1
5 m
pump
zero PE level
P = ? W = 1230 J t = 30 s
P =
P = 41 W
We assume the force produced by the pump is constant

9. Example 3.
A car, of mass 1000kg, produces 50 kW of power. The resistances to motion are given by R = 100v where v is the speed of the car. Find the maximum speed the car can attain ascending (going up) a road inclined at 10º to the horizontal.
V ms-1
Normal reaction F
P = W
R = 100v
1000g
10º
Use P = Fv
F =
Resolve forces parallel to the plane
F – R – 1000gsin10º = 0
- 1000v – = 0
50000 – 1000v² v = 0
Use the quadratic formula to solve
v = ms-1 (3 s.f.)

10. Try some yourself
Blue book pg 267, 268, 269, enjoy
Mechanics 2 pg 92 exercise 4c
all questions