Chapter 7 Kinetic Energy and Work

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Chapter 7Kinetic Energy and Work

• Kinetic Energy of a mass m is
• Energy is a scalar quantity
• The SI unit of energy is
• This unit is named after an English scientist of
  the 1800s, James Prescott Joule.

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Sample Problem 7-1In 1896 in Waco, Texas,
William Crush of the Katy railroad parked two
locomotives at opposite ends of a 6.4-km-long
track, fired them up, tied their throttles open,
and then allowed them to crash head-on at full
speed in front of 30,000 spectators.

• Hundreds of people were hurt by flying debris
  several were killed. Assuming each locomotive
  weighed 1.2 x 106 N and its acceleration along
  the track was a constant 0.26 m/s2, what was the
  total kinetic energy of the two locomotives just
  before the collision?

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SOLUTION 
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Work

• Work W is positive if kinetic energy is
  transferred to an object by an external force.
• Work is negative if kinetic energy is transferred
  from an object. In other words, negative work is
  done if an objects kinetic energy decreases.
• Work is a scalar and the SI unit is Joule.

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Work done by an External Force

• To calculate the work done on an object by a
  force during a displacement, we use only the
  force component along the object's displacement.
  The force component perpendicular to the
  displacement does zero work.

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• A force does positive work when it has a vector
  component in the same direction as the
  displacement, and it does negative work when it
  has a vector component in the opposite direction.
  It does zero work when it has no such vector
  component.

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Sample Problem 7-2Figure 7-4a shows two
industrial spies sliding an initially stationary
225 kg floor safe a displacement of
magnitude 8.50 m, straight toward their truck.

• The push of Spy 001 is 12.0 N, directed at
  an angle of 30 downward from the horizontal the
  pull of Spy 002 is 10.0 N, directed at 40
  above the horizontal. The magnitudes and
  directions of these forces do not change as the
  safe moves, and the floor and safe make
  frictionless contact.

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(a)  What is the net work done on the safe by
forces and during the displacement ?
SOLUTION 
Work done by
Work done by
Total work done
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(b)  During the displacement, what is the work Wg
done on the safe by the gravitational force
and what is the work WN done on the safe by the
normal force from the floor?
SOLUTION 

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(c)  The safe is initially stationary. What is
its speed vf at the end of the 8.50 m
displacement?
SOLUTION 
Work done on object equals increase in kinetic
energy
We have assumed no frictional forces exist.
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Sample Problem 7-3

• During a storm, a crate of crepe is sliding
  across a slick, oily parking lot through a
  displacement while a steady wind pushes
  against the crate with a force
• . The
  situation and coordinate axes are shown in Fig.
  7-5.

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(a)  How much work does this force from the wind
do on the crate during the displacement?
SOLUTION 
Work done by the wind force on crate
The wind force does negative work, i.e. kinetic
energy is taken out of the crate.
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(b)  If the crate has a kinetic energy of 10 J at
the beginning of displacement , what is its
kinetic energy at the end of ?
SOLUTION 
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Work done by Gravitational Force

• The angle ? is between (down) and .
  For the upward path, .
• The work done by the gravitational force is
  negative for the upward path.
• The change of gravitational potential energy of
  the object equals the negative of the work done
  by the gravitational force.

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• After the object has reached its maximum height
  and is falling back down, the angle ? 0o. The
  work done is

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Work done on Lifting and Lowering an Object
against a Field at Constant Velocity

• An applied force of magnitude mg pointed up can
  barely lift an object of weight mg. Assuming no
  change in the velocity, the work done by the
  applied force on the object in lifting it a
  distance d is
• The work done by the applied force in lifting the
  object is positive. It has the opposite sign as
  the work done by the gravitational force.

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• The applied force of magnitude mg (pointed up)
  lowers the object with no change in velocity. The
  work done by the applied force on the object in
  lowering it a distance d is
• The work done by the applied force in lowering
  the object is negative.
• The work done by the applied force equals the
  change in the gravitational potential energy of
  the object.

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Sample Problem 7-4Let us return to the lifting
feats of Andrey Chemerkin and Paul Anderson.

• (a)  Chemerkin made his record-breaking lift with
  rigidly connected objects (a barbell and disk
  weights) having a total mass m 260.0 kg he
  lifted them a distance of 2.0 m. During the lift,
  how much work was done on the objects by the
  gravitational force acting on them?

SOLUTION 
The gravitational force points down and the
displacement d points up
The work done by the gravitational force is
negative to the work done by the applied force.
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(b)  How much work was done on the objects by
Chemerkin's force during the lift?
SOLUTION 
The work done by the applied force is
(c)  While Chemerkin held the objects stationary
above his head, how much work was done on them by
his force?
Since the displacement d is zero, the work done
is zero.
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(d)  How much work was done by the force Paul
Anderson applied to lift objects with a total
weight of 27 900 N, a distance of 1.0 cm?
SOLUTION 
The work done by the applied force of Paul
Anderson is
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Sample Problem 7-5

• An initially stationary 15.0 kg crate of cheese
  wheels is pulled, via a cable, a distance L
  5.70 m up a frictionless ramp, to a height h of
  2.50 m, where it stops (Fig. 7-8a).
• (a)  How much work Wg is done on the crate by
  the gravitational force during the lift?

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SOLUTION 
The work done by the gravitational force during
the lift is negative
Since d sin ? h, we have
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(b)  How much work WT is done on the crate by the
force from the cable during the lift?
SOLUTION 
Since the crate has zero velocity before and
after the lift, the work done by the applied
force must be equal and opposite to the work done
by the gravitational force.
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Sample Problem 7-6

• An elevator cab of mass m 500 kg is descending
  with speed vi 4.0 m/s when its supporting cable
  begins to slip, allowing it to fall with constant
  acceleration /5.
• (a)  During the fall through a distance d 12 m,
  what is the work Wg done on the cab by the
  gravitational force ?

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SOLUTION 
During the fall, the work done on the cab by the
gravitational force is positive.
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(b)  During the 12 m fall, what is the work WT
done on the cab by the upward pull of the
elevator cable?
SOLUTION 
(positive direction is down)
This gives the magnitude of T, the direction of T
is up
The work done on the cab by T during the fall is
negative.
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(c)  What is the net work W done on the cab
during the fall?
SOLUTION 
(d)  What is the cab's kinetic energy at the end
of the 12 m fall?
SOLUTION 
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The Spring Force

• When a spring of length L0 is either compressed
  or extended, a restoring force acts opposite to
  .
• This restoring force is known as the spring force
  and is given by
• The constant k is a scalar and is called the
  spring constant or force constant. The SI unit is
  N/m.
• The negative sign indicates that the spring force
  is always opposite in direction to the
  displacement.
• Hookes law is named after Robert Hooke of the
  late 1600s.

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Work done by the Spring Force

• The work done by the spring force as the block is
  moved from position xi to xf is
• We have assumed that all the mass is at the block
  and the spring itself is massless.

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• The work done by the spring force on the block is
  negative if xf gt xi.
• If we set at L0 , xi 0, then the work done by
  the spring force on the block for extension x is
• Change in potential energy of spring - work
  done by the spring force.
• The work done by the spring force is to increase
  P.E. and to decrease K.E.

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Work Done on the Spring byan Applied Force

• If a block is stationary before and after a
  displacement, the work done on the block by an
  applied force is negative to the work done on it
  by the spring force.
• Therefore, for an extension x, the work done by
  an applied force on the block is
• The work done by the applied force equals to
  elastic potential energy of the spring.
• The work done by the applied force is to increase
  the P.E. of the spring.

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Sample Problem 7-7
A package of spicy Cajun pralines lies on a
frictionless floor, attached to the free end of a
spring in the arrangement of Fig. 7-10a. An
applied force of magnitude Fa 4.9 N would be
needed to hold the package stationary at x1 12
mm. (a)  How much work does the spring force do
on the package if the package is pulled rightward
from x0 0 to x2 17 mm?
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SOLUTION 
Work done by the spring is
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(b)  Next, the package is moved leftward to x3
-12 mm. How much work does the spring force do on
the package during this displacement? Explain the
sign of this work.
SOLUTION 
The spring force does positive work as the block
moves from 17mm to its relaxed position and
negative work as the block moves from the relaxed
position to -12mm. The former work is larger
resulting in WS being positive.
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Sample Problem 7-8

• In Fig. 7-11, a cumin canister of mass m 0.40
  kg slides across a horizontal frictionless
  counter with speed v 0.50 m/s. It then runs
  into and compresses a spring of spring constant k
  750 N/m. When the canister is momentarily
  stopped by the spring, by what distance d is the
  spring compressed?

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SOLUTION 
We assume the spring is massless. Work done by
the spring on the canister is negative. This work
is
Kinetic energy change of the canister is
Therefore,
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Work done by a General Variable Force
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Power

• The instantaneous power P is
• Power is a scalar and the SI unit is J / s which
  is known as the Watt (W), named after James Watt.
  
• Note that the kilowatt-hr is a unit of work

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The angle ? is between the force and velocity.

• Therefore,

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Sample Problem 7-10

• Figure 7-14 shows constant forces and
  acting on a box as the box slides rightward
  across a frictionless floor. Force is
  horizontal, with magnitude 2.0 N force is
  angled upward by 60 to the floor and has
  magnitude 4.0 N. The speed v of the box at a
  certain instant is 3.0 m/s.

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(a)  What is the power due to each force acting
on the box at that instant, and what is the net
power? Is the net power changing at that instant?
SOLUTION 
The kinetic energy of the box is not changing.
The speed of the box remains at 3 m/s. The net
power does not change.
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(b)  If the magnitude of is, instead, 6.0 N,
what now is the net power, and is it changing?
SOLUTION 
There is a net rate of transfer of energy to the
box. The kinetic energy of the box increases. The
net power also increases.
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Homework (due Oct 18)

• 5P
• 7E
• 11P
• 17P
• 19P
• 21E
• 23P
• 31E
• 33P