Transcript Document

AS Mechanics
Unit 2 (PHYA2)
NI
Newton’s Laws
• Objects stay at rest or in uniform motion (velocity,
i.e., speed + direction) unless acted on by an
unbalanced force.
AR
constant v
W
Notes:
• Add up the forces in each perpendicular direction (x, y, z) and
see if there is a resultant – i.e., non-zero force Fx, Fy or Fz.
• The Ancient Greeks believed bodies’ usual state was to be
still. They didn’t understand friction. Newton & Galileo
showed that moving bodies stay moving. Think satellites and
airtracks…
NII
Newton’s Laws
• The force on a body is proportional to
its rate of change of momentum.
Notes:
• We have defined the newton (N) so that ∑F = ∆p/∆t
• Since p = mv, ∑F = ∆mv/∆t we have two cases
(a) m constant: ∑F = m∆v/∆t = ma
(b) v constant: ∑F = v∆m/∆t
• Examples: space rocket, tractor pulling trailer, water through
hole in reservoir wall, winch. Case (a) or (b)?
• This applies in each perpendicular direction (x, y, z), i.e., ∑Fx
= max,
∑Fy = vy∆m/∆t, etc., etc.
Newton’s Laws
• “Objects stay at rest or move with uniform velocity
unless acted on by a force”
– No forces
– Balanced forces (i.e., resultant = 0)
NI
• Bigger forces cause bigger accelerations, larger
masses get smaller accelerations, hence F = ma
– F is the RESULTANT force (F1 + F2 + ...)
– a is ALWAYS in the same direction as F (even if v isn’t)
• The rate of change of momentum of a body is
NII
proportional to the resultant force on it.
– Hence F ~ mv/t
– 1 N is defined so that it is ‘equal’ as well as ‘proportional’
NIII
Newton’s Laws
• When body A exerts a force on body B, then
B exerts an equal (magnitude) and opposite
(direction) a force on A.
Notes:
• FA–B = –FB–A
• 6 pairs to find here:
• Four conditions to be a NIII pair
(a) same type of force,
(b) same magnitude,
(c) opposite directions,
(d) acting on different objects.
Weight
• Weight is a force
• F = ma, W = mg
g:
• the gravitational field strength, or
• the acceleration due to gravity
• is independent of m
• is 9.81 Nkg–1 (or 9.81 ms–2) in London
Classic mechanics problems
• Lifting mass
Classic mechanics problems
• Man in a lift
– Force on lift = T – mg = ma
– If lift is stationary or travelling at
constant velocity, T = mg
Change in
motion
Direction of
travel
accelerating decelerating
Up
T>mg
T<mg
Down
T<mg
T<mg
Classic mechanics problems
• Masses over a frictionless pulley
(Atwood’s machine)
• Tension in rope T is same on both
sides
• m2g – T = m2a
• T – m1g = m1a
• (adding) m2g – m1g = m2a + m1a
• or
m2  m1
a
g
m2  m1
• Can think of this as both masses being
accelerated by a force equal to the
difference in their weights
T
T
Classic mechanics problems
• Inclined slope, no friction
• Force acting down slope is
mgsinq
• So a = gsinq
• Block slides down slope
• Note that although it takes
longer to reach the bottom
than if it had dropped
vertically, its speed at the
bottom is the same – can
you explain why?
Classic mechanics problems
• Inclined slope, with friction
(F)
• Force acting down slope is
mgsinq – F
• So mgsinq – F = ma
• This time the velocity at
the bottom of the slope will
be less than the frictionless
case.
• Try the questions on p.137
Terminal speed
• Speed or velocity as
we’re only interested in
the downwards direction
• Drag depends on the
object’s shape and
speed, and the viscosity
of the fluid
• Maximum acceleration is
when v = 0. Use the
gradient of the v–t graph.
Counter
force
Horizontal
motion
Driving
force
Driving force – provided by rider/engine
Counter force – air resistance and friction
• Driving force < counter force:
vehicle slows down
• Driving force = counter force:
vehicle moves at constant
velocity
• Driving force > counter force:
vehicle speeds up
Stopping distances
• stopping = thinking + braking
s = s1 + s2
s1 = vt0 (speed × reaction time)
s2 = v2/2a (from v2 = u2 + 2as) (or s2 =
WD/Fbraking)
• Thinking distance a v
• Braking distance a v2
Impact time
• Impact time is the duration of an impact
force
• Remember s  u  v
t
2
Average speed
2s
• so impact time is t 
uv
• acceleration is
v u
a
t
Impact example
• A 500 kg car crashes into a lamp post.
The car stops from a speed of 13 ms–1 in a
distance of 1 m.
2s
2
t
  0.15s
• t=?
u  v 13
v u
13
2
a



86
.
7
ms
 8.8 g
• a=?
t
0.15
• F=?
F  ma  500  86.7  43350N
Road safety
• Airbags inflate to
slow the
deceleration of
the head and
spread pressure
over a wider
area
• Seatbelts are
designed to
stretch, slowing
the deceleration
of the body
• Crumple zones
in vehicles
increase the
deceleration
time
• Padding inside
crash helmet
increases
deceleration
time
• Now try Qs on p. 145
Work
• When you expend energy to exert a force which
moves an object, you are doing “work”.
work done  Force x distance in the direction of the force
Joules  Newtons 
metres 
• Work done = energy transferred
Is any work
being done in
these cases?
Work questions
• How much work is done when a mass of 3
kg is lifted vertically through 6 m?
– Work = F × d = (3 × 10) × 6 = 180 J
• A hiker climbs a hill 300 m high. If she
weighs 500 N calculate the work she does
lifting her body to the top of the hill.
– Work = F × d = 500 × 300
= 150,000 J or 150 kJ
Motion and force in different
directions
• Work done = component of average force
in direction of motion × distance moved in
direction of force
• W = Fscosq
• If q = 90°, W = 0.
Force–distance graphs
• Area under a force–distance graph is work
done
Constant force
varying force
Force extending a spring
F = kx
• So work to extend to
L is area under graph
1
W   L  F
2
force
• The more you stretch it, the harder you
have to pull
• Hooke’s Law:
F
L
extension
Kinetic energy
• KE is the result of work being done.
• Imagine a constant force F accelerating a
mass m:
t=0, u=0
v
F
m
m
uv
1
v u v
s
t  vt, a 

2
2
t
t
mv
F  ma 
t
mv vt 1 2
W  Fs 
  mv
t
2 2
Kinetic Energy
• Kinetic energy (KE) is the
energy an object has due to
its motion.
• So an object has more KE if:
1 2
KE  mv
2
mass
– it has a greater mass, or
– it moves faster
• A lorry and a van both travel
at 15 ms–1. The lorry has a
mass of 2000 kg and the van
a mass of 1000 kg. What KE
does each have?
30 kJ
15 kJ
velocity
Kinetic Energy
10 m/s
•
Calculate the KE of
a 500 kg car
travelling at
a) 10 ms–1
b) 20 ms–1
•
•
25 kJ
20 m/s
100 kJ
So what effect does
doubling the speed
have on the kinetic
energy? It increases by a factor of 4
How much work does the engine have to do to
increase the speed from 10 to 20 ms–1? 75 kJ
Work required to increase KE
• Ideally, the work done on an object is all
transferred to the increase in KE.
– e.g. a 40 kg block of ice is pushed across a
smooth floor with a force of 100 N for 5 m.
what is its final velocity?
Work done (F × d) = change in KE (½mv2)
so v2 = 2Fd/m = 2×100×5 / 40
So v = 5 m/s
• Note that this assumes no energy is
lost to friction or air resistance – in real
life this is never the case and the final
velocity will be smaller.
Potential Energy
• Potential energy (PE) is the energy
stored in an object when you raise it up
against the force of gravity.
• Energy change = work done
So PE gained = work done raising object
PE change = force × distance
=(mg) × h
• So change in PE = mgh
Potential energy
• A roofer carries 20 kg of
tiles up a 10 m ladder to
the roof.
– What is the gain in PE of
the tiles? 2 kJ
– How much work did the
roofer do lifting them? 2 kJ
– He accidentally drops one 500 g tile. How much
PE does it lose as it falls to the ground? 50 J
Conservation of energy
• “Energy cannot be created or destroyed, it
can only be transferred from one form to
another”
• What energy
transfers are chemical →
kinetic →potential
potential → kinetic
taking place in
the picture?
Objects falling due to gravity
• When an object falls, PE is converted to KE.
– e.g. a 20 kg cannon ball is dropped from the top
of the Eiffel tower, which is 320 m high. What is
the maximum speed of the ball as it hits the
ground?
loss of PE (mgh) = gain in KE (½mv2)
So v2 = 2mgh/m = 2×20×9.81×320 / 20
so v = 80 ms–1 (1sf)
• Why will the actual final speed be less than this?
– Wind resistance will limit it
PE and KE can interchange
• E.g. on a rollercoaster this can happen
several times…
In fact, work done to overcome friction
and air resistance = mgh – ½mv2
loss of PE (mgh) = gain in KE (½mv2)
v2 = 2mgh/m = 2gh = 2×9.81×50
Qs on p. 152
so v = 31.6 m/s
Power
• Power is the rate of transfer of energy
– (or the rate of doing work, therefore)
 E W
P

t
t
units: watt (W)
1 W = 1 Js–1
Motive power
• When a powered vehicle moves at
constant speed:
• Work done/s = force × distance/s = Fv
• Constant speed so resistive forces = motive force
• When a powered vehicle gains speed
• Motive power = gain in KE/s + energy lost to
resistance/s
• Accelerating so motive force > resistive forces
Efficiency
• No machine which converts energy from
one form to another is 100% efficient
– Some energy is always “lost”
– This is often due to friction of some kind
– “wasted” energy (all energy?) tends to end up
as heat
useful energy transferred by machine
Efficiency 
total energy supplied to machine
work done by machine
output power


energy supplied to machine input power
• Now try questions on pp.160–1