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MTH1150
Work and Energy
Definition
Work:
• A force is said to do work if, when acting on a mass there is a displacement of the
point of application in the direction of the force. Work is measured in Joules.
Energy:
• Energy is a property of objects which can be transferred to other objects or
converted into different forms. Energy is measured in Joules.
• Note that the SI unit for both work and energy is the Joule.
Joules
The Joule, symbol J, is a unit of energy. It is equal to the energy
transferred (or work done) to an object when a force of one newton
acts on that object in the direction of its motion through a distance of
one meter.
J=
𝑘𝑔∙𝑚2
𝑠2
= 𝑁 ∙ 𝑚𝑒𝑡𝑒𝑟𝑠
Work
Work can be described using the following formula:
𝑊 = 𝐹(𝑋𝑓 − 𝑋𝑖 )
Where:
• W is work measured in Joules
• F is force measured in Newtons
• (Xf-Xi) is change in position measured in meters
Work
• Work is a scalar quantity (as it is the product of two vectors), and is
determined relative to the direction of the motion.
• If there is no displacement, there is no work done.
• When we have broken a problem into x- and y-components, the work
becomes the sum of forces times the displacement in each direction.
• Wx = Σ Fx (Xfx - Xix)
• Wy = Σ Fy (Xfy - Xiy)
• Work done in the negative direction has a negative value, and
similarly, work done in the positive direction has a positive value.
Example 1
A person pulls his wagon 13
meters with a constant velocity
by applying a force of 200 N on
the handle at an angle of 65
degrees from the horizontal.
What is the net work done on
the wagon?
Example 1
The work done in the y-direction will be zero, because the wagon has
no displacement in this direction. The direction of the displacement
will be in the x-direction.
Example 1
• Wx = Σ Fx (Xfx - Xix)
• W = F cos(angle) (Xfx)
• W = 200 cos(65) * (13)
• W = 1099 J
• The net work applied to the wagon is 1098 J in the direction of the
displacement.
Example 2
A box is being pulled up a ramp
at a constant velocity. The
coefficient of kinetic friction is
0.75. What is the work done by
the force pulling the box up the
ramp? Would this work be
different if we were to have
lifted the box straight up to the
final height?
Example 2
The direction of the displacement is up the ramp.
• Sum Fy = 0
• Fn - W cos(angle) = 0
• Fn = 600 cos(35) = 491.5 N
Example 2
Sum Fx = 0 (constant velocity)
• F - W sin(angle) - μk*Fn = 0
• F = 0.75(491.5) + 600 sin(35)
• F = 712.8 N
Example 2
Calculate Work
• Wx = Σ Fx (Xfx - Xix)
• W = 712.8 (11)
• W = 7840 J
• The work done pulling the box up the ramp was 7840 J.
Example 2
• If one were to lift the box straight up (rather than using the ramp),
the height (displacement upward) would be 11*sin(35) = 6.3 m
• W = 600N * 6.3m = 3786 J
Example 2
It would only take 3786 J of work to lift the box. The difference is the
work lost to friction when the box is being pulled up the ramp. If there
was no friction, the work done by the lifting force would be the same
for each method.
Definition of Power
Power is the rate of doing work.
𝑊
𝑃=
𝑡
Where:
• P is power measured in Watts
• W is work measured in Joules
• t is time measured in seconds
Kinetic Energy
Kinetic Energy may be thought of as the energy embodied in a moving
object. It may be thought of as the Work required to bring a moving
object to rest.
Kinetic Energy
Kinetic Energy:
𝐾𝐸 =
1
𝑚𝑣 2
2
Where:
• KE is Kinetic Energy measured in Joules
• m is the mass of the object in kg
• v is the velocity of the object in m/s
The Work-Energy Theorem
• Work can be calculated based on the change in Kinetic energy. This is
called the Work-Energy Theorem.
The Work-Energy Theorem
1
1
1
2
2
𝑁𝑒𝑡 𝑊𝑜𝑟𝑘 = 𝐾𝐸𝑓 − 𝐾𝐸𝑖 = 𝑚𝑣𝑓 − 𝑚𝑣𝑖 = 𝑚(𝑣𝑓 2 − 𝑣𝑖 2 )
2
2
2
In other words, if there is no change in velocity, there is no net force
applied, and therefore, there is no work being done.
Potential Energy
Potential Energy is the energy embodied by an object due to its elevation above the
ground. You might think of it as the Work required to raise an object a certain
elevation.
𝑃𝐸 = 𝑚 ∙ 𝑔 ∙ ℎ
Where:
•
•
•
•
PE is potential energy measured in Joules
m is the objects mass measured in kg
g is the acceleration due to gravity
H is the objects height measured in meters
Net Work
𝑁𝑒𝑡 𝑊𝑜𝑟𝑘 = 𝑃𝐸𝑓 − 𝑃𝐸𝑖 = 𝑚 ∙ 𝑔 ∙ 𝑋𝑓𝑦 − 𝑚 ∙ 𝑔 ∙ 𝑋𝑖𝑦 = 𝑚𝑔(𝑋𝑓𝑦 − 𝑋𝑖𝑦 )
Where:
• PEf is the final potential energy in Joules
• PEi is the initial potential energy in Joules
• 𝑋𝑓𝑦 is the objects final height in meters
• 𝑋𝑖𝑦 is the objects initial height in meters
Combining Energy Equations
• The net work measured by a change in Kinetic Energy and the net
work measured by a change in Potential Energy may be combined
into a single formula.
• 𝑁𝑒𝑡 𝑊𝑜𝑟𝑘 = 𝑚𝑔 𝑋𝑓𝑦 − 𝑋𝑖𝑦 +
1
𝑚
2
𝑣𝑓 2 − 𝑣𝑖 2
• If the Net Work is positive, there is a net force adding energy to the
system. W = Force x Displacement.
• If the Net Work is negative, there is a net force removing energy from
the system. W = Force x Displacement.
Conservation of Energy
• If there is no energy added or lost to a system (no net work is done),
we can say that the energy embodied by an object at an initial
position, will be the same as the energy embodied by an object at a
final position. In other words, energy is conserved.
Conservation of Energy
• 𝐸𝑓 = 𝐸𝑖
• 𝑃𝐸𝑖 + 𝐾𝐸𝑖 = 𝑃𝐸𝑓 + 𝐾𝐸𝑓
• Or, if we substitute our formulae for Kinetic Energy and Potential
Energy:
𝑚𝑔 𝑋𝑓𝑦 − 𝑋𝑖𝑦
1
+ 𝑚 𝑣𝑓 2 − 𝑣𝑖 2 = 0
2
Example 1 – Conservation of Energy
A tennis player hits a ball (0.08 kg) that leaves her racket with a velocity
of 28 m/s at an angle of 40 degrees. The ball hits an empty seat in the
stands 5.5 m above the point where it was hit. What was the speed of
the ball hitting the empty seat?
Example 1 – Conservation of Energy
• The energy of the ball should be conserved, because once the ball has been
hit, there is no energy being added or removed (ignoring drag) from the
system.
• Using Conservation of Energy:
• mg(Xfy - Xiy) + 1/2m(Vf2 - Vi2) = 0
• 0.08 (9.81) (5.5 - 0) + 1/2 (0.08) (Vf2 - 282) = 0
• Vf = 26 m/s
Example 2 - Work
A snow-boarder (53 kg) is
sliding down a hill with a
slope of 38 degrees, and
after 40 m of vertical drop,
she has increased her
speed from 5 m/s to 25
m/s. What is the net work
done on the boarder?
Example 2 - Work
• Net Work = mg(Xfy - Xiy) + 1/2m(Vf2 - Vi2)
• Net Work = 53 (9.81) (-40) + 1/2 (53) (252 - 52) = -4897.2 J
• There is almost 4900 J lost to the system (likely due to friction).
• We would be able to confirm this using kinematics.
Example 2 - Work
Step 1 – Sum the forces in y to find Fn
• Sum Fy = 0
• Fn - W cos(38) = 0
• Fn = 53 (9.81) cos(38) = 409.7 N
Example 2 - Work
Step 2 – Find acceleration using Kinematics
• Xix = 0
Xfx = 40 / sin(38) = 65 m
Vix = 5 m/s
Vfx = 25 m/s
ax = ax
t=t
• (25 + 5) / 2 = (65 - 0) / t
• t = 4.33 s
• ax = (25 - 5) / 4.33 = 4.62 m/s2
Example 2 - Work
Step 3 – Sum the forces in x to find Fk
• W sin(38) - Fk = m*ax
• 53 (9.81) sin(38) - Fk = 53 (4.62)
• Fk = 75.5 N
Example 2 - Work
Step 4 – Calculate the work done by friction
• The work done by friction = Fk * Displacement
• Work = 75.5 * 65 = 4907 J
• The friction removed 4900 J of energy from the system. You should
notice that this is the same as the value determined through the
energy equation.