Transcript Physics_AS_Unit2_20_Power

What can you remember from P3 in Y11?
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Definition
Formula
Derived Units
Actual units
1. To understand how to successfully
complete mechanical power problems
2. To apply these skills to the slightly more
involved questions at AS
Book Reference : Pages 153-154
Some further thoughts on work done from
last lesson....
Two people each shifting boxes up a flight of
stairs... One at a full sprint, while the other
takes all day. Currently, through our view of
work done, we would calculate the work done
by each to be the same
Clearly this conflicts with our everyday
definition of “working hard”
Definition :
Power is defined as the rate of energy transfer
Formula :
Power = Energy
Time
And when energy is transferred by a force doing work...
Power =
Work done
time taken to do that work
Units:
Derived : Js-1 = called Watts (capital W)
Worked Example:
A person with a mass of 48kg (480N) climbs a flight
of stairs with a height of 10m in 12s
Power =
Work done
time taken to do that work
Power =
Power = 400W
480N x 10m
12s
Looking at the power Equation again:
Power =
Power =
Work done
time taken to do that work
Force x distance moved etc...
t
However, d/t is a very familiar concept.....
Power =
Force x velocity
Engines produce motive power. A powered
vehicle can be found in different scenarios:
1. Moving at constant speed & height
2. Moving & gaining speed
3. Moving & gaining height
No reason why we couldn’t mix and match!
Moving at constant speed & height
All of the resistive forces, (friction, air
resistance etc) are equal and opposite to the
motive force.
The work done by the engine is lost to the
surroundings (heat, sound etc)
PowerEngine = ForceResistive x velocity
Moving & gaining speed
The motive force from the engine exceeds
the resistive forces. We have an unbalanced
force and so we accelerate
The work done by the engine is the sum of
the energy lost to surrounding and the gain in
kinetic energy due to the increase in speed
PowerEngine = ForceResistive x velocity + K.E gain per sec
Note K.E. = ½mv2 coming soon.....
Moving & gaining height
If we are driving up an incline we are gaining
height.... If we are gaining height we are
gaining gravitational potential energy (GPE)
The work done by the engine is the sum of
the energy lost to the surroundings and the
gain in gravitational potential energy
PowerEngine = ForceResistive x velocity + GPE gain per sec
Note G.P.E. = mgh coming soon.....
Show that a juggernaut lorry with an output
power of 264kW moving at a constant speed
of 70 mph (31 m/s). What are the resistive
forces experienced?
PowerEngine = ForceResistive x velocity
ForceResistive = PowerEngine / velocity
ForceResistive = 264,000 / 31
ForceResistive = 8516N
Power is the rate of energy transfer and can be
calculated using:
Power =
Work done
time taken to do that work
When powered motion is involved we can use:
Power =
Force x velocity
This can be applied to scenarios with either
constant level velocity or where there are gains in
kinetic energy and/or potential energy due to
increases in velocity and height respectively.