Transcript Lesson 1 Introducing Newtons Second Law

Quick Starter
The blocks in the diagram below are in equilibrium, g = 10ms-2
Find the friction force on the 4kg block and the tensions in the ropes.
4 kg
6
kg
3
kg
Lesson Objectives:
1) Be able to recognise when a system is not in
equilibrium
2) Begin to use Newton’s second law to find missing
forces when an object is not in equilibrium
Newton’s first law:
A particle/body is in equilibrium if it is at rest or travelling at
constant velocity in a straight line
For each of the following situations decide if ‘Fred
Lemming’ is in equilibrium or not.
‘Fred Lemming’ is standing on the floor and not moving
‘Fred Lemming’ is sitting in a car that is slowing down as
some traffic lights approach
‘Fred Lemming’ is sitting at rest on a table that is sloping at
30 degrees to the horizontal
‘Fred Lemming’ is standing in a lift as it starts to move up
three floors
‘Fred Lemming’ is sitting in a car that is travelling along a
straight road at 30 mph
‘Fred Lemming’ is sitting on a ‘Merry-Go-Round’ as it spins
round at the fun fair
‘Fred Lemming’ is falling under gravity
Newton’s second law:
A body that is not in equilibrium must be accelerating/decelerating
The Resultant Force causing the acceleration is found using the
formula:
Resultant Force = Mass of the body × Acceleration
OR
Forces in Direction of Motion – Forces in Opposite Direction =
Mass of the body × Acceleration
Forces in Direction of Motion – Forces in Opposite Direction = Mass
of the body × Acceleration
Eg. The 2kg block below is moving to the right.
Write an expression for the Resultant Force and find the
acceleration.
3N
2N
2 Kg
7N
Forces in Direction of Motion – Forces in Opposite Direction = Mass
of the body × Acceleration
Eg. The 3kg block below is moving to the left.
Write an expression for the Resultant Force and find the
acceleration.
4N
1N
3 kg
8N
In the pack are 10 cards:
Five are force diagrams
Five are situation descriptors
Match them up and label the forces on the diagrams!
When you have finished pick a card. Find the acceleration
of the object on the card and use it to find the Resistance
force present in the question.
Note 1: The difficulty of the questions increase with the
card. Card A is the easiest and E is the hardest.
Note 2: Do not start with E!
A
A mass of 0.5kg is at rest but
falls into a pool of water 3m
deep. It takes 4 seconds to
reach the bottom.
C
D
A car of mass 2000kg is
travelling at 10ms-2 along a
flat road when it brakes with a
constant deceleration so that it
can stop at some traffic lights
25m away. The braking force
of the car is 3700N.
2
A rocket of mass 7,000kg
starts from rest and accelerates
to 30ms-1 in the time it takes to
travel 50m. The driving force
of the rocket is 135000N.
3
4
B
A car of mass 2000kg driving
along a flat road accelerates
from 5ms-1 to 10ms-1 in 30
seconds. The driving force of
the car is 3500N.
1
A
A mass of 0.5kg is at rest but
falls into a pool of water 3m
deep. It takes 4 seconds to
reach the bottom.
v2 = u2 + 2as
s = ut + ½at2
v = u + at
B
A car of mass 2000kg driving
along a flat road accelerates
from 5ms-1 to 10ms-1 in 30
seconds. The driving force of
the car is 3500N.
v2 = u2 + 2as
s = ut + ½at2
v = u + at
C
A rocket of mass 7,000kg
starts from rest and accelerates
to 30ms-1 in the time it takes to
travel 50m. The driving force
of the rocket is 135000N.
v2 = u2 + 2as
s = ut + ½at2
v = u + at
D
A car of mass 2000kg is
travelling at 10ms-2 along a
flat road when it brakes with a
constant deceleration so that it
can stop at some traffic lights
25m away. The braking force
of the car is 3700N.
v2 = u2 + 2as
s = ut + ½at2
v = u + at
E
A car of mass 1000kg tows a
caravan of mass 750kg along a
horizontal road. The engine of
the car exerts a force of
2500N. The resistance force
on the caravan is half of the
resistance force on the car.
The vehicles are travelling at
5ms-1 and accelerate to 7ms-1
in 12m.
v2 = u2 + 2as
s = ut + ½at2
v = u + at