1. Mechanical Work
• Mechanical work done by a force (F) is the force’s
Work, Power, & Energy magnitude times the displacement of an object in
the direction of the force
Objectives: W = Fx ∆px + Fy ∆py
• Define and learn to compute work, power,
where:
kinetic energy, potential energy, strain
energy – Fx , Fy : x- and y-components of force
• Understand and apply the principle of – ∆px , ∆py : x- and y-components of displacement
conservation of energy • Units of Work:
• Understand and apply the principle of work – English: foot-pound (ft·lb) = (1 ft)(1 lb)
and energy
– SI: Joule (J) = (1 N)(1 m)
Positive vs. Negative Work Example Problem #1
• Positive work : results when force and displacement
At the start of a bobsled run, a bobsledder pushes
are in the same direction. Indicates that the force is
on the sled with a constant force of 70 N, at an
acting to speed up the object angle of 30° below the forward direction.
• Negative work: results when force and displacement
The bobsledder stops pushing after the sled has
are in opposite directions. Indicates that the force is
moved 30 m forward.
acting to slow down the object
How much work has the bobsledder done on the
If bicycle moves sled?
Fbrake Fpedal
displacement ∆px:
Fpedal does positive
work
Fbrake does negative
∆px work
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2. Power Instantaneous Power
• Power = the rate of work production • Instantaneous power represents the rate at which
work is being performed at a specific instant.
• Power calculated as work done in a given time:
• Instantaneous power is calculated as force times
the velocity in the direction of the force:
Work W
Power = P=
Change in Time ∆t
P = Fx vx + Fy vy
• Power can be generated (P > 0) or absorbed (P < 0)
Since:
W F x ∆px + Fy ∆py
• Units of Power: P= =
– English: horsepower (hp) = 550 ft·lb/s ∆t ∆t
– SI: Watt (W) = (1 J)/(1 s) F x ∆px + F y ∆py
=
∆t ∆t
Example Problem #2 Kinetic Energy
A weightlifter lifts a 200 kg mass from the ground to • Energy = Capacity to do work
a height of 2 m in 1 s. • Kinetic Energy (KE) = the energy associated with
He holds the mass in the air for 3 s. the motion of an object
He lowers the mass back to the ground in 2 s. • KE equals the amount of work the object can do in
being brought to rest
Assume that the work performed by the weightlifter
is the opposite of the work performed by gravity. 1
What is the work performed by the weightlifter
KE = 2 m v²
during each phase of the lift? where:
What is the total work performed? – m : mass of the object
What is the average power generated/absorbed by – v : magnitude of the object’s velocity
the weightlifter during each phase?
• Units of KE = Units of Work (Joule, foot-pound)
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3. Potential Energy Strain Energy
• Potential energy (PE) = stored energy that can be • Strain Energy (or Elastic Energy) is potential energy
converted into work due to the deformation of an object
• Associated with an object’s position/configuration • Is the work that can be done through a return to the
• PE due to gravity = the weight (W) of an object original shape
times its height (h) above a reference height
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PEse = k x2
PEg = W h = –mgh 2 x
m : mass of the object where:
W h k : spring constant
g : acceleration due to gravity
(stiffness of the object)
(–9.81 m/s2)
x : deformation from
• Units of PE = Units of KE, Work unloaded length
Conservation of Energy Example Problem #3
• When gravity is the only external force acting on a When a 70 kg diver leaves the 2 m board, her
system, the total energy of the system remains center of mass has a projection speed of 10 m/s,
constant: a projection angle of 60°, and a projection height
of 3 m above the water.
KE + PE = constant Make a rough sketch of kinetic energy, potential
energy, and total energy vs. time
1 1 What was the maximum height of her center of
2
mv12 - mgh1 = 2
mv22 - mgh2 mass during flight? (use the principle of
conservation of energy)
• Projectile motion is one case when conservation of
energy applies.
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4. Principle of Work and Energy Example Problem #4
• The work produced by an external force is equal to During a cheer, an 85 kg male cheerleader has to
the change in the total energy of the object acted catch a 45 kg female cheerleader.
upon:
At the time that the man starts to catch her, the
woman’s center of mass is 2 m above the
W = ∆KE + ∆PE + ∆TE ground and is falling downward at 4 m/s
At the end of the catch, the woman’s center of
where: mass is 1 m above the ground.
– W : work
How much work did the male cheerleader do
– ∆KE : change in kinetic energy (KEfinal – KEintial) during the catch?
– ∆PE : change in potential energy (PEfinal – PEintial)
What average vertical force did he need to exert?
– ∆TE : change in thermal energy (TEfinal – TEintial)
Friction Static Friction
• Shear force acting over the area of contact between • Static friction is equal in magnitude and opposite in
two bodies, and in a direction that opposes motion direction to the sum of all other shear forces
• Two types of friction: • Their exists a maximum static friction force:
– Static Friction : applies when the objects are not
sliding relative to each other Ffriction ≤ µs N ΣFshear
– Kinetic Friction : applies when the objects are Ffriction
sliding relative to each other where:
µs : coefficient of static friction N
v (a property of the two materials
in contact; 0 ≤ µs ≤ 1)
Ffriction N : normal force between surfaces
• If required static friction force exceeds maximum
static friction force, object will begin to slide
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5. Kinetic Friction
• Magnitude of the kinetic friction force is proportional
to the normal force between the two surfaces:
v
Ffriction = µk N
Ffriction
where:
µk : coefficient of kinetic friction
N
(a property of the two materials
in contact; 0 ≤ µk < µs )
N : normal force between surfaces
• Direction of the kinetic friction force is always
opposite to the direction of sliding
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