1. Force and motion Objectives When you have competed it, you should
* understands Newton’s first law of motion
* know some different types of forces
* know and be able to apply Newton’s second law to simple examples of objects moving in a straight line
* understands the idea of equilibrium.

2. Force and motion Newton’s first law
A particle will remain at rest or will continue to move with constant velocity in a straight line unless the forces act on it to change that state.
Example
When you are riding a bicycle on a level path and start to free-wheel, you can keep up an almost constant velocity force some time.
But eventually you will slow down, partly because of air resistance.
direction of motion
air resistance.

3. Force and motion Newton’s first law Example
When a stone is sent sliding across the frozen surface of a lake. However smooth the ice appears, the stone eventually slows down because of friction between the two surfaces

4. Force and motion Force = mass  acceleration F = ma
Newton’ second law
The force F applied to a article is proportional to the mass m of the particle and the acceleration produced.
Force = mass  acceleration
F = ma
SI units:
Mass 
kilogram (kg)
Acceleration 
metres per second per second (m s-2)
Force 
Newton (N)
1 N is the force needed to give 1 kg an acceleration of 1 ms-2.

5. Types of forces
Pushing
or Thrust T
Tension
D N
Driving force

6. Force and motion F = m a F = 400  1.2 = 480 N Example
An ice-yacht of mass 400 kg has an acceleration of 1.2 ms-2. What force is needed to produce this ?
Solution
1.2 ms-2
400 kg F
F = m a
F = 400  1.2 =
480 N

7. Force and motion Example
Find the magnitude of the acceleration produced when a particle of mass 5 kg is acted upon by a force of magnitude 22 N.
Solution
a ms-2
5 kg
F = 22 N
F = ma
22 = 5  a
a =
4.4 ms-2

8. Force and motion Example
A car of mass 1000 kg is pushed with a force of 200 N. Calculate the acceleration of the ca; and find how long it will take to reach a speed of 3 ms-1 form rest.
Solution
200 = 1000  a
: a = 0.2 ms-2
F = m a :
Data: u = 0 v = a = 0.2 t?
Required equation v = u + at

9. Force and motion Example
A curling stone of mass 18 kg is launched across ice with a speed of 2 ms-1, and goes a distance of 25 metres before coming to rest. Calculate the deceleration, and find frictional ore between the stone and the ice.
Solution
Data:
u = 2
v =
s = 25
: 02 =  a  25
Use the equation :
v2 = u2 + 2as
gives: a =
ms-2
F = ma :
F = 18  ( ) = N