Download Unit 6 Mechanical Principles and Applications

Unit 5 Mechanical Principles
and Applications
M2 Work and Energy Part 1
Re-cap on P4
Remember:
v=u+at
s=ut+1/2at2
v2=u2+2as
s=1/2(u+v)t
where:
s=displacement, u=initial velocity, v=final
velocity, a=linear acceleration.
Work
When a force F moves a distance x, work done,W,
is given by: W=Fx
Types of Energy
In the first diagram a mass is raised on a
pulley. Work is done and the energy of the
mass increases. This energy is stored in
the mass as potential energy.
In the second diagram a truck is accelerated
along a surface. A force is needed to do
this and work is done. The energy used to
accelerate the truck is stored as kinetic
energy.
Potential Energy
This is the energy
stored in a mass as it
is moved upwards
against the force of
gravity. The weight is
mg, the distance
moved is h metres
(sometimes z or x).
W = mgh
Worked example
If a mass being lifted is 200 kg and it is
raised 0.6 m, determine the work done
and the change in P.E. of the mass.
Weight = mg, so the force to be overcome =
200 x 9.81 = 1962 N
W=F x = 1962 x 0.6
= 1177.2 J
Kinetic Energy
This is the energy stored in a mass due to its
velocity.
F = ma
a = v/t
x = vt/2
W = fx
We can re-arrange all these to get:
K.E. = ½ mv2
Worked example
A force of 80 N is used to pull a truck 200 m
along a horizontal floor. Determine the
work done and the increase in K.E.
W = Fx = 80 x 200 = 16000 J
If no energy is lost due to friction then this
must all end up as the Kinetic Energy of
the truck.
Conservation of Energy
The law of conservation of energy states
that energy cannot be created or
destroyed, but changed from one form in
to another.
In many real situations friction causes
energy to be “lost” as heat. In the following
questions we will assume that there are no
losses due to friction.
Falling Bodies
A body, which is z metres above a point, has
a P.E. of mgz. As it falls P.E. is converted
into K.E., ½ mv2.
mgz = ½ mv2
2mgz = mv2
v2 = 2gz
v = √(2gz)
Worked Example
A ball of mass 0.4 kg swings on the end of a
tin rod with negligible mass with length
60 mm. The ball is held horizontal and
then released. Calculate the following:
i. the velocity of the ball as it passes
through its lowest position.
ii. the loss of potential energy.
iii. the gain in kinetic energy.
Answer
i. the ball will swing through a vertical height
of 60 mm.
v = √(2gz) =√(2 x 9.81 x 0.06) = 1.085 m/s.
ii. change in P.E. = mgz = 0.4 x 9.81 x 0.06
= 0.2354 J
iii. change in K.E. = ½ mv2 = ½ x 0.4 x
1.0852 = 0.2354 J
Note energy is conserved.
Question 1
An object of mass 20 kg is dropped onto a
surface from a height of 50 m. Calculate
the energy and velocity just before it hits
the surface.
Question 2
A swinging hammer must have 50 Joules of
energy and a velocity of 2 m/s at the
bottom of the swing. Calculate the mass
and height of the hammer before it is
released.
Question 3
A swinging hammer has a mass of 2 kg and
is raised 0.2 m. calculate the energy and
velocity at the bottom of the swing.