Transcript Work Energy Powerpoint

Physics
Work and Kinetic Energy
Teacher: Luiz Izola
Chapter Preview
1. Work Done by a Constant Force
2. Kinetic Energy
3. Work-Energy Theorem
4. Work Done by A variable Force
5. Power
Introduction
 We know by intuition that motion,
energy, and work are somehow related.
 Work can be thought as force times
displacement.
 In this chapter will show the relation
between work and energy of motion
(kinetic energy).
Work Done by a Constant Force
 First, we will consider the case where
force and displacement have the same
direction.
 Later, we will consider cases where force
and displacement have arbitrary directions.
 Then, we will learn how to calculate work
done on an object which has several forces
acting on it.
Force in the Direction of Displacement
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The greater the force, the greater the work.
The greater the distance, the greater the work.
Work = Force x Displacement
Measured in Joules (J) (Newton x Meter)
Work is a scalar. (Not a vector).
1J = kg. m2/s2
Force in the Direction of Displacement
Ex: A finch can exert a force of 205N with its beak
as it cracks a seed case. If the beak moves
0.40cm, how much work does it do to get the
seed?
Ex: An intern pushes a 72-kg patient on a 15-kg
gurney, producing an acceleration of 0.60m/s2.
How much work does the intern do by pushing the
patient 2.50meters. Assume no friction.
Force at an Angle to Displacement
 A person pulling a suitcase on a level surface
makes an angle Ө with the horizontal.
Work is the component of force in the direction
of the displacement times the displacement.
W = (FcosӨ)d = Fdcos(Ө). The unit is J.
What is the effect on work for Ө = 0o or 90o?
Force at an Angle to Displacement
 Work is the component of force in the direction of
displacement times the magnitude of the
displacement or the component of displacement
in the direction of force times the magnitude of
the force.
 Remember that the result is a scalar. It is just the
multiplication of the magnitudes of displacement
and force vectors.
Force at an Angle to Displacement
Ex: A 75-kg person slides 5.0-m on a water slide,
dropping through a vertical line of 2.50-m. How
much work does the gravity do on the person?
Force at an Angle to Displacement
Ex: You want to load a box into a truck. One way is to
lift the box straight up through a height “h”, doing a
work W1. Otherwise, you can slide the box up a ramp
a distance “L”, doing work W2. Which is correct: (a) W1
< W2 (b) W1 = W2 (c) W1 > W2
Negative Work and Total Work
 Work depends on the angle between force and
displacement (or motion direction).
• Work is positive if the force has a component in
the direction of motion (Ө < 90o).
• Work is zero if the force has no component in the
direction of motion (Ө = 90o).
• Work is negative if the force has a component in
the direction of motion (Ө > 90o).
When more than one force acts on an object, the
total work is the sum of the work done by each force
separately or the work of the Ftotal.
Negative Work and Total Work
Ex: A car of mass “m” coasts down a hill inclined at
an angle Ф below the horizontal. The car is
acted on by 3 forces: (a) The normal force N
applied by the road, (b) The force due to air
resistance Fair, and (c) The gravity force mg.
Find the total work done on the car as it travels a
distance “d” along the road.
Negative Work and Total Work
Ex: Calculate the total work from the previous example by
using Wtotal = FtotaldcosФ.
Kinetic Energy and Work-Energy Theorem
 If the total work done in an object is positive,
its speed increases.
 If the total work done in an object is
negative, its speed decreases.
 Knowing that Kinetic Energy, K, is defined
as:
K = 1/2mv2,
where: m = mass
v = velocity.
The unit is (kg)(m2/s2) = J (Joule)
Kinetic Energy and Work-Energy Theorem
Ex: A truck moving at 15m/s has a kinetic energy
of 1.4x105 J. What is the mass of the truck?
Work-Energy Theorem
“The total work done on an object is equal to
the change in its kinetic energy.”
Wtotal = ΔK = 1/2mvf2 – 1/2mvi2
Ex: How much work is required for a 74-kg sprinter
to accelerate from rest to 2.2m/s?
Kinetic Energy and Work-Energy Theorem
Ex: A 4.1-kg box of books is lifted vertically from rest
a distance of 1.60m by an upward force of 60.0N.
Find (a) Work done by the applied force. (b) The
work done by gravity. (c) The final speed of the box.
Kinetic Energy and Work-Energy Theorem
Ex: A boy exerts a force of 11.0N at 29.0o above the
horizontal on a 6.40-kg sled. Find the work done
by the boy and the final speed of the sled as it
moves 2.00m, assuming the sled starts with an
initial speed of 0.500m/s and slides horizontally
without friction.
Work Done by a Variable Force
 If a force varies continuously with position, we
can approximate it with a series of constant
values that follow the shape of a curve.
 The work done by the continuous force is
approximately equal to the sum of the rectangles’
areas.
 In the limit of an infinite number of very small
rectangles, the total area of all rectangles
becomes identical to the area under the curve.
Work Done by a Variable Force
Work Done by a Variable Force
 A particular case is the work to stretch/compress a
spring a distance x from equilibrium
W = 1/2kx2 (k measures in N/m)
Work Done by a Variable Force
Ex: (a) If the work required to stretch the Slinky
Dog 1 meter is 2J, what is the force constant for
the Slinky Dog? (b) How much work is required
to stretch it from 1m to 2m?
Power
 Power is a measure of how quick the work is
done.
 Therefore, power is defined as:
P = Work (W) / Time (t) = W/t = Fv
 The International Unit is: (watts) W = J/s
Ex: To pass a truck, your 1.30x103kg car needs to
accelerate from 13.4m/s to 17.9m/s in 3.00s.
What is the minimum power required to pass
it?
Power
Ex: It takes a force of 1280N to keep a 1500-kg car
moving with constant speed up to a slope of
5.0o. If the engine delivers 50 hp (50x746W) to
the drive wheels, what is the maximum speed
of the car?
Ex: Calculate the power output of a 1.3g fly as it
walks straight up a window pane at 2.5cm/s.
Homework
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