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Engineering Mechanics 2
Course
:
Diploma in Digital & Precision Engineering (EGDF10)
Diploma in Aeronautical & Aerospace Technology (EGDF11)
Diploma in Engineering with Business (EGDF17)
Module Title :
Engineering Mechanics 2
Module Code :
EGD203 / EGF203 / EGM203
Topic 4:
Kinetics of Particles – Work and Energy
4.1
Introduction
4.2
Work of a force
4.3
Work and kinetic energy
4.4
Work of a spring force
4.5
Conservative forces and potential energy
4.6
Conservation of energy
4.7
Power and mechanical efficiency
OBJECTIVES:
At the end of this lesson, the students should be able to:
• Differentiate kinetic energy and potential energy
• Use energy equation to perform kinematic calculations
Topic 4
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Topic 4
4.1
Kinetics of Particles – Work and Energy
Introduction
In the study of dynamics, other than forces, sometimes we also like to analyse the position and
velocity of a body (or particle). The concept of work and energy provides a tool to help us to
perform this task.
The work-energy method combines the principles of kinematics with Newton’s second law to
directly relate the displacement (position) and velocity of a body. As such, it is not a new or
independent principle. There are no problems that can be solved with the work-energy method
that cannot be solved using Newton’s second law. However, when the work-energy does apply,
it is usually the easiest method of solving a problem.
4.2
Work of a force
In mechanics, a force does work only when the particles to which the force is applied
moves. For example, when a constant force F is applied to a particle, which moves a
displacement s in a straight line as shown in Fig 4.1, the work done on the particle by the force
F is defined by the scalar product (see Topic 1) of the force and the displacement.
Work done, U = F . s (in plain English, Work done is force times displacement)
U
=F.s
= F s cosθ (Scalar product of 2 vectors)
Fig 4.1 – Work done by F to move a mass through displacement s
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Work may be interpreted as either:
•
the product of F and the component of displacement in the direction of F, i.e., s cosθ
U = F ( s cos θ )
•
(4.1)
the product of s and the component of force in the direction of s, i.e., F cosθ
U = s ( F cos θ )
(4.2)
Note the Eqn (4.1) and (4.2) are the same mathematically. In a more sophisticated way of
defining work done is:
S2
Work done, U = ∫ F.ds
S1
(Note: The result of scalar product is a scalar)
The above equation allows us to handle cases where the force F can vary along a general path
(may not be straight). Work has dimensions of force × length. In the SI system of units, this
combination of dimensions is called Joule (1 J = 1 N.m)
Recall: Dot product of 2 vectors (result is not a vector, but a scalar)
If a and b are 2 vectors with magnitude a and b and directions as shown, then the dot product
of these 2 vectors will produce a scalar quantity with magnitude equals to {ab cosθ} where θ
is the angle between vectors a and b
θ
a
4.3
a.b = ab cosθ
b
Work and kinetic energy
Fig 4.2 – force moving along a general path from s 1 to s 2
Topic 4
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As shown in Fig 4.2, the work done, U 1-2 , by a force along a path from s 1 to s 2 is:
s2
s2
s1
s1
U 1− 2 = ∫ F ⋅ ds = ∫ F cos θ ds (F and θ are variables that changes along s)
We are now ready to discuss the principle of work and energy.
Considering motion in the tangential direction,
∑ F cos θ = ma
Note: Work done in the normal direction
is always zero as cos90° = 0.
t
⇒ ∑ F cos θ
= m (dv/dt)
⇒ ∑ F cos θ
= m (dv/ds) (ds/dt)
⇒ ∑ F cos θ
= m v (dv/ds)
(chain rule)
⇒ ∑ F cosθ ds = m v dv
Integrating along the path from s 1 to s 2 with speeds v 1 to v 2 :
∑ ∫s F cos θds = ∫v mvdv
or
s2
v2
1
1
1
1
∑ ∫s F cos θds = 2 mv2 2 − 2 mv12
s2
1
(4.3)
If we define the kinetic energy of a particle to be
T=
1 2
mv
2
(4.4)
Then from equations (4.3) and (4.4), we get
∑U
1− 2
= T2 − T1
T1 + ∑ U 1− 2 = T2
Or
(4.5)
Eqn (4.5) is a scalar equation. Thus, only one unknown can be obtained by using this equation.
In particular, it cannot be used to determine forces directed normal to the path of motion because
such forces do no work! For a system of particles, the principle is applied to each particle in the
system and the scalar equations are added algebraically, so that
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∑U
1− 2
= ∑ T2 − ∑ T1
(4.6)
Note that
•
The principle of work and energy is used to solve kinetic problems that involve velocity,
force and displacement.
•
Kinetic energy is energy of motion. The kinetic energy of an object is the energy it
possesses because of its motion (velocity).
Procedure of analysis
The procedure of analysis is as followed:
•
draw the free body diagram
•
apply principle of work and energy
•
apply kinematics / equation of motion if required
Example 4.1
A 50-kg crate moves 5 m down a 15° chute with an initial velocity v 1 of 6 m/s at position ○,1 as
E
A
shown. Determine the velocity v 2 when the crate reaches position ○,2. Assume the coefficient
E
R
R
A
A
of kinetic friction µ = 0.35.
Solution:
First, we draw the free-body diagram as shown below:
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Note: The FBD remains the same
from position (1) to (2). In general,
this may not be true. If such case
happens, FBD will need to be drawn
at various stages.
Next, we realize that the work done by the component of the weight down the plane is positive
(displacement and force in same direction), whereas that done by the friction force is negative
(displacement and force in opposite direction).
From statics, in the y direction, Σ Fy =0
⇒ R = mg cos 15°
⇒ R = (50) (9.81) cos 15°
⇒ R = 474 N
The total work done by all forces acting on the crate during the motion is
ΣU 1-2 = Work done by R + Work done by friction + Work done by weight
R
R
ΣU 1-2 = 0 + (-µ R)(5) + (mg) (5 sin 15°)
R
R
= – (0.35) (474) (5) + (50) (9.81) (5 sin 15°)
= – 194.7 J
Note: Work done by R is always zero because its direction is always perpendicular to the
displacement. Also, since the net work done is negative, we obtain a decrease in the
kinetic energy.
The kinetic energy:
T 1 = ½ m(v 1 )2
R
Topic 4
R
R
R
P
⇒
T 1 = ½ (50)(6)2
⇒
T 1 = 900 J
R
R
R
P
R
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Engineering Mechanics 2
T 2 = ½ m(v 2 )2
R
⇒
R
R
R
P
T 2 = ½ (50)( v 2 )2
R
R
R
R
P
The work-energy equation (4.5) gives
900 -194.7 = ½ (50)( v 2 )2
R
⇒
R
P
= 5.31 m/s
v2
Example 4.2
A 1200 kg sports car travels down the 5° inclined road at a speed of 10 m/s as shown below.
If the driver applies the brakes causing all wheels to lock, determine the distance s that the tyres
will skid on the road. The co-efficient of kinetic friction between the wheels and the road is
given as µ k = 0.4
R
R
Solution:
This problem can be solved using the principle of work and energy since it involves force,
velocity, and displacement. Force and displacement enable us to calculate work done, whereas
velocity enables the kinetic energy to be calculated.
Free-Body diagram:
Note: As stated earlier, normal force N does zero work since it never undergoes displacement
along its line of action
+
Topic 4
Σ Fn = 0
R
R
(n- direction, perpendicular to the slope)
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⇒ N − mg cos 5° = 0
⇒ N = (1200) (9.81) cos 5°
= 11727 Newton
Friction Force = µ k N
R
R
= (0.4) (11727)
= 4691 Newton
Principle of Work and Energy
∑U
= T2 − T1
1− 2
mg sin 5° (s) − µ k N (s) =
R
R
(
1
m v 2 2 − v12
2
)
1
{(1200) (9.81) sin5° − 4691}s = ( )(1200)(0 2 − 10 2 )
2
⇒
3665 s = 60000
⇒
s = 16.37 m
Alternate Solution to Example 4.3
An alternate solution is to use equation of motion. First, from the free body diagram, the
equation of motion is applied along the inclined surface to determine the acceleration.
Σ F x = ma x
+
R
R
R
R
(x direction)
⇒ mg sin 5° − µ k N = ma
R
R
⇒ (1200) (9.81) sin5° − 4691 = 1200 a
⇒ a = − 3.054 m/s
Next, since the acceleration is a constant, we can use one of the constant acceleration kinematic
equations to solve for distance s.
v2 = v o 2 + 2a (s)
P
P
R
RP
P
02 = 102 + 2 (-3.054) (s)
P
Topic 4
P
P
P
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∴ s
= 100/6.108
= 16.37 m
4.4
Work of a spring force
The magnitude of a force developed in a linear elastic spring when the spring is displaced a
distance s from its un-stretched position is F = k.s, where k is the spring stiffness. As shown in
Fig.4.3, the spring is elongated or compressed from a position s 1 to a further position s 2 under a
R
R
R
R
spring force F s
R
(a)
(b)
Fig 4.3 – Work done by a spring force during (a) extension (b) compression
The work done on the spring by the force Fs is positive, since in each case the force and
displacement are in the same direction. We have:
U1-2
s2
= ∫ Fs ds
s1
s
= ∫ 2 ks ds
s1
=
1 2 1 2
ks 2 − ks1
2
2
(4.7)
However, if a particle is attached to a spring, then the force Fs exerted on the particle is opposite
to that exerted on the spring as shown in Fig.4.4.
Topic 4
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Engineering Mechanics 2
Direction of displacement
Fig.4.4 – Spring force exerted on a particle when it is stretched
Consequently, the force will do negative work on the particle when the particle is moving so as
to further elongate (or compress) the spring. Hence the above equation becomes
U1-2
1
1
= − ( ks 22 − ks12 )
2
2
(4.8)
Tips: Mistake in sign can be eliminated when one simply notes the direction of the spring acting
on the particle and compares it with the direction of displacement of the particles –
a) if both are in the same direction, positive work results, and
b) if they are opposite to one another, the work is negative.
4.5
Conservative forces and potential energy
In this section and section 4.6, we try to introduce another way to view the same problems
above. Instead of analyse using kinetic energy and work done, we use the concept of energy, in
terms of potential energy and kinetic energy. You will realize that potential energy is actually
similar to work done.
A force is called a conservative force if it is:

dependent only on the position of the particle,

independent of the velocity and acceleration of the particle, or
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Engineering Mechanics 2

if the work done is independent of the path followed by the particle.
We state here without proof, that the weight of a particle and the force of an elastic spring are
examples of conservative forces. When a conservative force acts on a particle, the force has the
capacity to do work. This capacity, called the potential energy, depends only on the position of
the particle when acted upon by the force.
Examples include the gravitational potential energy and the elastic potential energy. The
potential energy can often be interpreted as the external work done in order to bring the particle
from its datum position to the current position.
4.6
Conservation of energy
In solving many problems, friction forces are often negligible and the only forces acting on the particle
are elastics springs and gravity.
When a particle is acted upon by a system of conservative force, the following equations hold:
T1 + V1 = T2 + V2
(4.9)
1
where Ti = Kinetic energy ( mv 2 ) at position i
2
1
Vi = gravitational potential energy (mgh) or elastic potential energy ( ks 2 ) at position i.
2
That is, when the only forces acting on a particle are conservative forces, the total mechanical
energy remains constant.
E1 = E2 = constant
Eqn (4.9) is often referred to as the Principle of Conservation of Energy.
The conservation of energy theorem is used to solve kinetics problems that involve velocity,
displacement and conservative force systems.
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In fact, equation (4.9) and equations (4.5), (4.6) are exactly the same; it is merely looking the
same thing from different angle of view. Notice that the potential energy terms in (4.9) are
lumped into the work done term in (4.5).
Energy as a tool for mechanics problem solving
Principle of an object free falling from rest (which means initial zero velocity) is illustrated in
Fig.4.5.
Fig.4.5 – Free falling object
Procedure for analysis
•
Potential and kinetic energy
•
Conservation of energy theorem
•
Kinematics/ Equation of motion (if required)
The procedure is illustrated by the following examples.
Example 4.3
A 0.8 kg ball is drop from rest at 20 m above the floor. Find its velocities when it is 12 m above
the floor and when it just hits the floor.
Solution:
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Take reference from the floor, where the P.E (or V) is equate to zero
VA = (0.8)g(20)
and
TA = 0
VB = (0.8)g(12)
and
TB = ½ (0.8)vB2
P
Where
Vi denotes the potential energy (P.E) of the ball at position i
Ti denotes the kinetic energy (K.E) of the ball at position i
By conservation of total energy at positions A and B,
TA + VA = TB + VB
0 + 0.8g(20) = ½ (0.8)vB2 + (0.8)g(12)
P
P
(0.8)(8)g = ½ (0.8)vB2
P
vB = 12.53 m/s
P
P
Similarly, this time we may consider position A and position C, where the ball just hits the
ground.
TA + VA = TC + VC
0 + 0.8g(20) = ½ (0.8)vC2 + 0
P
P
40g = vB2
P
vC = 19.81 m/s
P
P
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Example 4.4
The 10 kg cylinder is released from rest when the spring is un-stretched as shown. The spring
stiffness is 200 N/m. Determine the maximum extension x of the spring from its initial rest
position.
Solution:
Use conservation of energy equation (from moment cylinder released from spring till it just
reach its maximum travel)
T1 + V1 = T2 + V2 ————(a)
At position A,
KE
TA
=
1 2
mv A
2
1
= × 10 × 0 2
2
=0 J
PE
VA
= mgh
= (10) (9.81) x
= 98.1 x
At position B ,
KE
TB
=
1 2
mv B
2
= ½ (10)(0)2
P
=0 J
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Engineering Mechanics 2
PE
VB
= mgh
=0 J
Spring Energy absorbed
=
1 2
kx
2
= ½ (200) x2
P
= 100 x2
P
P
Putting all above into Eqn (a) gives
98.1x + 0 = 0 + 100 x 2
where x = total spring deformation
100 x 2 − 98.1x + 0 = 0
x2 – 0.981 x =0
P
P
∴ x = 0.981 m
or
x = 0 (trivial)
This problem can also be solved by using work and energy equation (4.5):
Free body diagram:
From the free body diagram, it shows that there are 2 forces; mg
and FSpring acting on the block. Each of them will contribute to
the ∑UA-B term.
Work done by mg = mgx
Work done by FSpring = – ½ k x2
P
TA + ∑UA-B = TB
⇒
½ mvA2 + [ mgx + (- ½ k x2) ] = 0
⇒
½ (10) (0)2 + [ (10)(9.81)x – ½ (200) x2 ] = 0
⇒
100 x2 – 98.1 x = 0
⇒
x = 0 (trivial)
Topic 4
P
P
P
P
P
P
P
P
P
P
or
x = 0.981 m
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Engineering Mechanics 2
Example 4.5
A 10 kg block is released from rest on the horizontal surface at point B, where the spring has
been stretched a distance of 0.5 m from the neutral position A. The co-efficient of kinetic friction
between the block and the plane is µk = 0.30. The spring constant Ks = 300 N/m. Determine the
velocity v of the block as it passes point A; and the maximum distance x to the left of A that the
block goes (at C where the block stops momentarily).
Solution:
Given: m =10 kg, ks = 300 N/m , x =0.5 m, vC = 0, vB = 0, µk=0.3
Use Work & Energy equation at position B→ A
TA + ∑UA-B = TB
—————– (a)
At position B, the kinetic energy TB:
TB
=
1 2
mv B
2
=0J
(since the block is released from rest at B, velocity at B is zero)
At position A, the kinetic energy TA :
TA
=
1 2
mv A
2
1
2
= ( )(10)v A
2
= 5v A 2
Work done ∑ U B − A comprised work done by frictional force (Ufriction) & spring force (Uspring)
Ufriction = Scalar product of (frictional force) and (displacement)
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= − ( µ mg) (0.5) (-ve as the frictional force always in opposite direction with displacement)
= − (0.3) (10) (9.81) (0.5)
= − 14.715 J
Uspring =
1 2
kx
2
(formula derived earlier)
= (1/2) (300) (0.5)2
= 37.5 J (positive because now the spring is pushing the block)
Putting all above into Eqn (a) gives
5v A 2 + {− 14.715 + 37.5} = 0
∴ vA = 2.135 m/s
Similarly, we now apply Work & Energy equation to position A→ C
TA + ∑UA-C = TC
—————– (b)
At position A, kinetic energy of the block is:
TA
=
1 2
mv A
2
= ½ (10) (2.135)2
= 22.68 J
At position C, kinetic energy of the block is:
TC
=
1 2
mvC
2
=1/2 (60) (0)2
=0 J
Work done by friction:
Ufriction = − (0.3) (10) (9.81) x
= − 29.43 x (Notice that work done by friction is always negative)
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Engineering Mechanics 2
1
Work done by spring Uspring = − kx 2
2
= -1/2 (300) x2
= − 150 x2 (negative because the force is opposing the displacement)
Putting all above into Eqn (b) gives
22.68 + {− 29.43 x − 150 x2} = 0
x2 + 0.196 x − 0.152 = 0
∴ x = 0.304 m
or
x = -0.5 m (reject)
Example 4.6
A cylindrical ram of mass 100 kg is released from rest 0.75m from the top of spring A with
stiffness of 12 kN/m. Nested in spring A is a second spring B of stiffness 15 kN/m. The
arrangement and un-stretched lengths of the springs are shown in the figure. Determine the
deflection of spring A needed to stop the ram’s downward motion. Neglect the mass of the
springs.
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Solution:
Assume ram compresses both springs at the instant it comes to rest.
Let spring A be compressed a distance = xA
So spring B is compressed by a distance = xB = (xA –0.1)
Also, the ram has thus dropped a total distance = (xA + 0.75)
Use Conservation of energy equation
T1 + V1 = T2 + V2
—————-(a)
where T1 = T2 = 0
V1 = mg(xA + 0.75)
1
1
2
V2 = { k A x A + k B ( x A − 0.1) 2 }
2
2
Hence Eqn (a) becomes
0 + mg (xA + 0.75)
1
1
= 0 + { k A x A 2 + k B ( x A − 0.1) 2 )
2
2
100 (9.81) (xA+0.75) =
981xA + 735.7
1
1
(12000) x A 2 + (15000)( x A − 0.1) 2
2
2
= 6000 (xA)2 + 7500 (xA − 0.1) (xA − 0.1)
13500 (xA)2 − 2481 xA − 660.7 = 0
Solving for xA,
xA = 0.331, –0.148 (discard the –ve value as it does not represent the physical condition)
Since xA must be positive, the solution is therefore
and
Topic 4
xA
= 0.331 m
xB
= (xA −0.1)
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Engineering Mechanics 2
= 0.231 m
which justifies the assumption that the block compresses both springs.
Note that you can also use the work and energy equation to solve this problem.
Example 4.7
A block of mass 2 kg is given an initial velocity of 1 m/s when it is at the top surface of the
smooth cylinder as shown. If the block moves along a path of radius 0.5 m, find the angle θmax at
which it begins to leave the cylinder’s surface.
Solution:
Free-body diagram
W = mg = (2) (9.81) = 19.62 N
Height of mass h = 0.5 (1-cos θmax)
Key note is that at θmax where the block just leaves the curve surface, the reaction from the
surface is zero. Also, there are only 2 forces acting on the block from ○,1 to ○,2 and the work
E
E
A
A
A
done by N is always zero, the only work done is by W. Recall that work done by weight is
always mgh.
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Engineering Mechanics 2
Principle of work and energy
T1 + ∑U1-2 = T2
Let v2 be the velocity of the block when it leaves the surface at position ○,2, so
E
A
A
½ mv12 + mgh = ½ mv22
⇒
½ (2)(1)2 + 19.62(0.5)(1-cosθmax) = ½ (2) v22
⇒
v 2 = 1 + 9.81(1 − cos θ max )
2
———– (a)
Equation of motion
Consider motion in the normal direction and apply the equation
∑ Fn = ma n
we get:
[where an =
v2
]
r
 v2 2 
− N + 19.62 cosθ = 2 

 0.5 
———– (b)
At θ = θmax , the block leaves the surface so that N = 0. Thus, Eqn (b) becomes
 v2 2 
− 0 + 19.62 cosθ max = 2 

 0.5 
v2 2
cos θ max =
4.905
———– (c)
Combine equations (a) and (c), we should get:
4.905 cos θ max = 1 + 9.81(1 − cos θ max )
4.905 cosθmax = 1 + 9.81 – 9.81 cosθmax
14.715 cosθmax = 10.81
∴ θmax = 42.7°
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Engineering Mechanics 2
4.7
Power and mechanical efficiency
Power and efficiency are concepts that are closely related to the concepts of mechanical work
and mechanical energy.
Power. The output power of a force F is given the rate at which it does work
Power =
dU F .dr
=
= F .v
dt
dt
Power has the dimension of work (force × length) divided by time. In the SI systems of units, it
is called watts (1 W = 1J/s=1 N.m/s)
Point to ponder: Is power a vector quantity?
Mechanical efficiency.
All mechanical systems lose energy to friction so that the amount of work output is always less
than the amount of work input.
The mechanical efficiency ηmech of the machine is defined by the ration of these two quantities.
η mech =
=
output work
input work
power output
power input
(4.10)
Since all real systems lose energy to friction, the output work is always less than the input work
and ηmech is always less than one.
Point to ponder: What is the unit of mechanical efficiency?
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Restricted and Non-Sensitive
EGF203 Engineering Mechanics
Project (Individual)
Deadline: 22ND January 2024 Sunday,
23:59
Aim:
The aim of this assignment is for you to perform a self-directed learning task which is outside the scope
of the class lectures. It is important to see why the mechanics we learn is useful in real-world
situations. Mechanics equations are a key component of many engineering problems. Through this
self-directed exploration you should be able to apply the work and energy principle which is useful for
designing real engineering problem.
Assignment:
Work and Energy equation or Conservation of energy equations are one of the most powerful
equations for solving different engineering applications. In class you are taught to solve
problems for mass or spring questions
In this assignment, you are expected to perform a research and self-directed study on a design
project.
Design a project to determine the specifications (unstretched length and spring constant k) for
the elastic cord to be used at a bungee-jumping facility. Participants are to jump from a platform
50m above the ground. When they rebound, they must avoid an obstacle that extends 4 m
below the point at which they jump.
In determining the specifications for the cord, establish reasonable the safety limits for the
minimum distances by which participants must avoid the ground and obstacle. Account for the
fact that participants will have different weight. (if needed, specify a maximum allowable weight
for participants)
Write a brief report (maximum 3 pages inclusive of pictures) including:
Restricted and Non-Sensitive
Your design report must have the following contents:
1.
2.
3.
4.
Design and specification considerations
Formulas and calculations
Discussion
Conclusion
•
Analysis – A brief design consideration on how to analyse the engineering project.
•
Specifications – Show the step-by-step calculation needed to design the specification of
the cord length and spring constant.
Design Recommendation – Clearly state all the assumptions you have made and recommend
a suitable specification of the cord.
•
The assignment report must be submitted in a single pdf file. The working can be typed or
handwritten format.
Grading:
This assignment is part of your assessment for this module and accounts for 20% of your overall
grade.
The scoring will be determined by the following:
1. Mathematical Analysis: Clear analysis on the design considerations
2. Report: The report is neatly presented and is submitted on time
3. Engineering discussion: The application of the how the Mechanics -Dynamics course has been
used and applied to this Engineering project assignment and assumptions and errors predicted
in using the principles.
4. Specification: Clear step-by-step worked solution showing how you solve the problem
statement on clear assumption stated.

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