Chapter 6 – Work and Energy
Old assignments and midterm exams (solutions have been posted on the web) can be picked up in my office (LB-212)
Assignment 6 Textbook (Giancoli, 6th edition), Chapter 6:
Due on Thursday, October 30, 2008
- Problem 19 - page 162 of the textbook - Problems 22 and 31 - page 163 of the textbook - Problem 89 - page 166 of the textbook
Chapter 6 (Today)
• Work Done by a Constant Force • Kinetic Energy, and the Work-Energy Principle • Potential Energy
Recalling Last Lecture
Work Done by a Constant Force
The work done by a constant force is defined as the distance moved multiplied by the component of the force in the direction of displacement:
(6-1)
Work Done by a Constant Force Problem 6-10 (textbook): What is the minimum work needed to push a 950-kg car 810 m up along a 9.0º incline? (a) Ignore friction. (b) Assume the effective coefficient of friction retarding the car is 0.25.
Work Done by a Constant Force Problem 6-10:
Draw a free-body diagram of the car on the incline. Include a frictional force, but ignore it in part (a) of the problem. The minimum work will occur when the car is moved at a constant velocity.
r F
r FP r F fr
y N
x
θ θ
r mg
Work Done by a Constant Force Problem 6-10:
(a) Write Newton’s 2nd law in both the x and y directions, noting that the car is unaccelerated
∑F ∑F
y
= FN − mg cos θ = 0 → FN = mg cos θ
x
= FP − mg sin θ = 0 → FP = mg sin θ
r Ther work done by FP in moving the car a distance d along the plane (parallel to FP ) is given by
(
WP = FP d cos 0o = mgd sin θ = ( 950 kg ) 9.80 m s 2
o 6 810 m sin 9.0 = 1.2 × 10 J ( ) )
Work Done by a Constant Force Problem 6-10:
(b)
Ffr = µ k FN
Now include the frictional force, given by:
We still assume that the car is not accelerated. We again write Newton’s 2nd law for each direction: The y-forces are unchanged by the addition of friction, and so we still have
FN = m g cos θ In the x-direction we have
∑F
= FP − Ffr − mg sin θ = 0 → FP = Ffr + mg sin θ = µ k mg cos θ + mg sin θ r r The work done by F P in moving the car a distance d along the plane (parallel to F P ) x
is given by W P = FP d cos 0 o = mgd ( sin θ + µ k cos θ
(
= ( 950 kg ) 9.80 m s 2
)
) ( 810 m ) ( sin 9.0
o
)
+ 0.25 cos 9.0 o = 3.0 × 10 6 J
Kinetic Energy, and the Work Energy Principle I mentioned, in the begin of this lecture, that the concept of energy is very important in physics. Yet, I started discussing about work done by a force. What is the connection between these two quantities? A good definition of energy was introduced by the Einstein’s theory of relativity: E = mc2 But this is beyond the scope of this course. In mechanics, we can use a less precise definition of energy as: “The ability to do work” In this course we will be defining translational kinetic energy and some forms of potential energies, though other forms the energy exist such as nuclear energy, heat energy, etc.
Kinetic Energy, and the Work Energy Principle Kinetic Energy A moving object can do work on a second object. Example: A car hits another car the first car applies a force on the second (say, at rest) which consequently undergoes a displacement. This has the implication that the first car has the ability to do work, consequently it has (or carries) energy.
The energy of motion is called kinetic energy (or KE for short).
But… how do we measure kinetic energy?
Kinetic Energy, and the Work Energy Principle How do we measure kinetic energy? Consider a rigid body of mass m, say the bus in the figure, moving along a line. A net force
acts on the bus changing its velocity from displacement of magnitude d.
The work done on the bus by the force Using Newton’s 2nd law,
to
over a
is given by:
, we can rewrite the above equation as:
Kinetic Energy, and the Work Energy Principle How do we measure kinetic energy? Previous slide: We can now use the equation of motion
with
x – x0 = d v = v2 v0 = v1
(6-2)
Kinetic Energy, and the Work Energy Principle How do we measure kinetic energy?
Previous slide:
We define translational kinetic energy as: (6-3) With this definition, we can rewrite eq. 6.2: (6-4) Where,
(6-5)
We can also write
(6-6)
where
(6-7)
Note: Eq. 6.3 is valid in more general cases even when the force varies.
Kinetic Energy, and the Work Energy Principle How do we measure kinetic energy?
Previous slide:
Equation 6.6 represents the work-energy principle which can be stated as
“ The net work done on a object is equal to the change in the object’s kinetic energy ”
The work-energy principle is ONLY valid for the net work done on an object: in other words, the work done by the net force applied on the object.
Kinetic Energy, and the Work Energy Principle Notes: Observing the previous equations:
We can conclude the following 1) If the net work done on an object is positive, then ∆KE > 0 implying that there was an increase in the object’s velocity ( v2 > v1 ). 2) If the net work done on an object is negative, then ∆KE < 0 implying that there was an decrease in the object’s velocity ( v2 < v1 ).
Kinetic Energy, and the Work Energy Principle Notes: Because work and kinetic energy can be equated, they must have the same units: kinetic energy is measured in joules. It is also a scalar quantity.
The change (decrease) in kinetic energy of the hammer is equal to the energy gained by the nail (it starts moving) or, more generally: The decrease in kinetic energy of the hammer is equal to the work the hammer can do on other objects.
Kinetic Energy, and the Work Energy Principle Problem 6-21 (textbook): If the speed of a car is increased by 50%, by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver’s reaction time. Assume: 1) Only friction acts along the direction of the car’s displacement 2) The driver locks the brakes.
Kinetic Energy, and the Work Energy Principle Problem 6-21: The work needed to stop the car is equal to the change in the car’s kinetic energy. That work comes from the force of friction on the car. Assume the maximum possible frictional force, which results in the minimum braking distance. Thus
F fr = µ s F N Where the static friction is used since the driver locks the brakes. The normal force is equal to the car’s weight if it is on a level surface, and so
F fr = µ s m g In the diagram, the car is traveling to the right.
d = stopping distance
r Ffr
r FN
r mg
W = ∆KE → Ffr d cos180o = 12 mv22 − 12 mv12 → − µ s mgd = − 12 mv12 → d = 2
v12 2g µs
Since d ∝ v 1 , if the velocity is increases by 50%, or is multiplied by 1.5, then d will be multiplied by a factor of (1.5)2 , or 2.25.
Work Done by a Constant Force Problem 6-25 (textbook): A 285-kg load is lifted 22.0 m vertically with an acceleration a = 0.160g by a single cable. Determine (a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the load, (d) the work done by gravity on the load, and (e) the final speed of the load assuming it started from rest.
Work Done by a Constant Force Problem 6-25: (a) From the free-body diagram for the load being lifted, write Newton’s 2nd law for the vertical direction, with up being positive. Fnet =
∑F
= FT − m g = m a = 0.160 m g
(
→
)
FT = 1.16 m g = 1.16 ( 285 kg ) 9.80 m s 2 = 3.24 × 10 3 N
r FT
r mg
Kinetic Energy, and the Work Energy Principle Problem 6-25: (b) The net work done on the load is found from the net force.
(
W net = Fnet d cos 0 o = ( 0.160 m g ) d = 0.160 ( 285 kg ) 9.80 m s 2 = 9.83 × 10 3 J
r FT
r mg
) ( 22.0 m )
Kinetic Energy, and the Work Energy Principle Problem 6-25: (c) The work done by the cable on the load is
(
Wcable = FT d cos 0 o = (1.160 mg ) d = 1.16 ( 285 kg ) 9.80 m s 2
r FT
r mg
) ( 22.0 m ) =
7.13 × 10 4 J
Kinetic Energy, and the Work Energy Principle Problem 6-25: (d) The work done by gravity on the load is
(
WG = mgd cos180 o = − mgd = − ( 285 kg ) 9.80 m s 2
r FT
r mg
) ( 22.0 m ) =
− 6.14 × 10 4 J
Kinetic Energy, and the Work Energy Principle Problem 6-25: (e) Use the work-energy theory to find the final speed, with an initial speed of 0.
W net = K E 2 − K E 1 = v2 =
2W n et m
+ v
2 1
=
1 2
m v 22 −
1 2
m v 12
(
2 9 .8 3 × 1 0 3 J 285 kg
r FT
r mg
→
)+0
= 8 .3 1 m s
