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Chapter 6
Work and Energy
6.1 Work Done by a Constant Force
FsW =( )J joule 1 mN 1 =!
Work involves force and displacement.
6.1 Work Done by a Constant Force
6.1 Work Done by a Constant Force
( )sFW !cos=
1180cos
090cos
10cos
!=
=
=
o
o
o
More general definition of the work done on an object, W, by a constant force, F, through a displacement, s, with an angle, θ, between F and s:
6.1 Work Done by a Constant Force
Example 1 Pulling a Suitcase-on-Wheels
Find the work done if the force is 45.0 N, the angle is 50.0 degrees, and the displacement is 75.0 m.
( ) ( )[ ]( )
J 2170
m 0.750.50cosN 0.45cos
=
== osFW !
6.1 Work Done by a Constant Force
( ) FssFW == 0cos
( ) FssFW !== 180cos
A weightlifter doing positive andnegative work on the barbell:
Lifting phase --> positive work.
Lowering phase --> negative work.
Example. Work done skateboarding. A skateboarder is coasting down aramp and there are three forces acting on her: her weight W (675 Nmagnitude) , a frictional force f (125 N magnitude) that opposes her motion,and a normal force FN (612 N magnitude). Determine the net work done bythe three forces when she coasts for a distance of 9.2 m.
W = (F cos θ)s , work done by each force
W = (675 cos 65o)(9.2) = 2620 J
W = (125 cos 180o)(9.2) = - 1150 J
W = (612 cos 90o)(9.2) = 0 J
The net work done by the three forces is: 2620 J - 1150 J +0 J = 1470 J
6.1 Work Done by a Constant Force
Example 3 Accelerating a Crate
The truck is accelerating ata rate of +1.50 m/s2. The massof the crate is 120 kg and itdoes not slip. The magnitude ofthe displacement is 65 m.
What is the total work done on the crate by all of the forces acting on it?
6.1 Work Done by a Constant Force
The angle between the displacementand the normal force is 90 degrees.
The angle between the displacement and the weight is also 90 degrees.
( ) 090cos == sFW
Thus, for FN and W,
6.1 Work Done by a Constant Force
The angle between the displacementand the friction force is 0 degrees.
( )[ ]( ) J102.1m 650cosN180 4!==W
( )( ) N180sm5.1kg 120 2 === mafs
6.2 The Work-Energy Theorem and Kinetic Energy
Consider a constant net external force acting on an object.
The object is displaced a distance s, in the same direction asthe net force.
The work is simply
s
!F
W = (Σ F)s = (ma)s
6.2 The Work-Energy Theorem and Kinetic Energy
( ) ( ) 2212
2122
21
ofof mvmvvvmasmW !=!==
( )axvv of 222 +=
( ) ( )2221
of vvax !=
DEFINITION OF KINETIC ENERGY
The kinetic energy KE of and object with mass mand speed v is given by
221KE mv=
constant accelerationkinematics formula
6.2 The Work-Energy Theorem and Kinetic Energy
THE WORK-ENERGY THEOREM
When a net external force does work on an object, the kineticenergy of the object changes according to
2212
f21
of KEKE omvmvW !=!=
(also valid for curved paths and non-constant accelerations)
6.2 The Work-Energy Theorem and Kinetic Energy
Example 4 Deep Space 1
The mass of the space probe is 474 kg and its initial velocityis 275 m/s. If the 56.0 mN force acts on the probe through adisplacement of 2.42×109m, what is its final speed?
6.2 The Work-Energy Theorem and Kinetic Energy
2212
f21W omvmv !=
W = [(Σ F)cos θ]s
6.2 The Work-Energy Theorem and Kinetic Energy
( ) ( ) ( ) ( )( )2212
f2192- sm275kg 474kg 474m1042.20cosN105.60 !="" vo
( )[ ] 2212
f21cosF omvmvs !=" #
sm805=fv
[(Σ F)cos θ]s
Example. A 58 kg skier is coasting down a 25o slope as shown. Near thetop of the slope, her speed is 3.6 m/s. She accelerates down the slopebecause of the gravitational force, even though a kinetic frictional forceof magnitude 71 N opposes her motion. Ignoring air resistance, determinethe speed at a point that is displaced 57 m downhill.
6.2 The Work-Energy Theorem and Kinetic Energy
6.2 The Work-Energy Theorem and Kinetic Energy
vo = 3.6 m/s s = 57 m 25o slope m = 58 kg fk = 71 N
Find vf
kfmgF !=" o25sin
6.2 The Work-Energy Theorem and Kinetic Energy
2212
f21W omvmv !=
W = [(Σ F)cos θ]s
= (58)(9.8)(0.42) - 71 = 170 N
W = [(170)cos 0](57) = 9700 J
Solving the Work-Energy Theorem for vf:
vf = [(2W + mvo2)/m]1/2 = [(2(9700) + (58)(3.6)2)/(58)]1/2
= 19 m/s
Work-Energy Theorem
6.2 The Work-Energy Theorem and Kinetic Energy
Conceptual Example 6 Work and Kinetic Energy
A satellite is moving about the earth in a circular orbit and an elliptical orbit. For these two orbits, determine whetherthe kinetic energy of the satellite changes during the motion.
6.3 Gravitational Potential Energy
( )sFW !cos=
( )fo hhmgW !=gravity
Find the work done by gravity inaccelerating the basketballdownward from ho to hf.
6.3 Gravitational Potential Energy
( )fo hhmgW !=gravity
We get the sameresult for Wgravityfor any path takenbetween ho and hf.
6.3 Gravitational Potential Energy
Example 7 A Gymnast on a Trampoline
The gymnast leaves the trampoline at an initial height of 1.20 mand reaches a maximum height of 4.80 m before falling back down. What was the initial speed of the gymnast?
6.3 Gravitational Potential Energy
2212
f21W omvmv !=
( )fo hhmgW !=gravity
( ) 221
ofo mvhhmg !=!
( )foo hhgv !!= 2
( )( ) sm40.8m 80.4m 20.1sm80.92 2 =!!=ov
6.3 Gravitational Potential Energy
fo mghmghW !=gravity
DEFINITION OF GRAVITATIONAL POTENTIAL ENERGY
The gravitational potential energy PE is the energy that anobject of mass m has by virtue of its position relative to thesurface of the earth. That position is measured by the heighth of the object relative to an arbitrary zero level:
mgh=PE
( )J joule 1 mN 1 =!
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