Download Work and kinetic energy

PHSX 114, Monday, September
29, 2003
• Reading for today: Chapter 6 (6-3 -- 6-7)
• Reading for next lecture (Wed.): Chapter 6
(6-7 -- 6-10)
• Homework for today's lecture: Chapter 6,
question 13; problems 22, 30, 39, 41
Work and kinetic energy
• Work W= F||d =Fd cosθ
• Kinetic energy (KE) is defined to be
KE=½mv2
• Units of both are in Joules
The work-kinetic energy
theorem
• Using calculus can prove Wnet=ΔKE
• Wnet is the total work done by all the forces
acting on the object
• ΔKE is final KE minus initial KE
• See p. 151 for proof in the special case of
constant forces
• Proof comes from Newton's second law and the
definitions of work and KE
• Example
d=5 m
Your turn
37o
A 1 kg block goes 5 m down a frictionless
incline that makes an angle of 37o with the
horizontal. If the block starts from rest,
what is its final speed?
Answer: As before, W = Fd cos(53o) =
(1 kg)(9.8 m/s2)(5 m) cos(53o)=29.5 J;
so 29.5 J = KE2-KE1, KE1=0, so
29.5 J=½mv2 => v2= 2(29.5 J)/(1 kg) =>
v=7.7 m/s
Potential energy (PE)
• Kinetic energy is associated with the object's
motion
• Potential energy is associated with
conservative external forces acting on the body
• Change in potential energy is minus the work
done by the force (ΔPE= - W)
• We'll discuss gravity and the elastic spring
force
Gravitational potential energy
• Raise ball from y1 to y2
• Work done by gravity is W = -mg(y2 - y1)
• The ball has gained gravitational potential
energy
• ΔPE= - W; ΔPE= mg(y2 - y1)
• PEgrav= mgy
• Example
Elastic spring force
• Hooke's Law: FS = -kx
• k is the spring constant
• x is the magnitude of the displacement from
the spring's rest position
• direction of the force is opposite the
direction of the displacement x
Elastic potential energy
• Force pulling the spring does work
WP=½kx2
• WP= -WS= ΔPE=½kx2
• PEelastic=½kx2
Example
A spring has constant k=500 N/m.
a) It is stretched from rest to x=0.02 m. Find the
PE, the work done to pull the spring and the
work done by the spring.
b) Continue stretching to x=0.04 m. Find the
additional work done to pull the spring and the
additional work done by the spring.
Example (cont.)
A spring has constant k=500 N/m.
c) Your turn: If the spring is compressed from
x1=0 to x2=-0.02 m, find the PE, the work done
to push the spring and the work done by the
spring.
Answer: W= ΔPE = PE2- PE1= ½kx22 - ½kx12=
½(500 N/m)(-0.02 m)2=0.10 J is work to push
spring; work done by spring is -ΔPE =-0.10J.
Conservative forces
• We cannot define PE for all forces
• To define a PE, work must depend on initial and
final positions and not the path in between
• Such a force is called "conservative"
• Gravity and the elastic spring force are
conservative forces
• Other conservative forces: electric force,
Newton's universal gravity
Non-conservative forces
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•
•
•
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The path taken matters to the work done
Cannot define a potential energy function
Mechanical energy is lost as heat
Examples: friction, air resistance
Compare work done by gravity, friction for
two different paths from height y
Conservation of mechanical energy
• Recall this example -• A 1 kg block goes 5 m down a frictionless incline
that makes an angle of 37o with the horizontal. If
the block starts from rest, what is its final speed?
• Previous answer: W=mgy=29.5 J; so
29.5 J = KE2-KE1, KE1=0, so 29.5 J = ½mv2 =>
v2= 2(29.5 J)/(1 kg) => v=7.7 m/s
• We want to rework this in the language of
conservation of energy
Conservation of energy approach
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•
•
•
Initially energy is all gravitational PE
As object falls PE is converted to KE
At the bottom, all the energy is kinetic
W = ΔKE = - ΔPE =>
(KE2 - KE1) = - (PE2 - PE1) =>
(KE1 + PE1) = (KE2 + PE2)
• E = KE + PE has the same value throughout
the motion
• E is "total mechanical energy"
Revisiting the 1 kg block problem
A 1 kg block goes 5 m down a
frictionless incline that makes an
angle of 37o with the horizontal.
If the block starts from rest, what
is its final speed?
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•
•
•
•
1
d=5 m
37o
v1= 0, so KE1 = 0
y1= 3 m, so PE1 = mgy1= (1 kg)(9.8 m/s2)(3 m) = 29.5 J
E = KE1 + PE1 = 29.5 J
y2= 0, so PE2 = 0
E = KE2 + PE2 = 29.5 J => KE2 = 29.5 J =½mv2 =>
v2= 2(29.5 J)/(1 kg) => v=7.7 m/s
2
Pendulum energy analysis
• If no energy lost to friction, pendulum
returns to same height
• Energy all potential at top of swing
• Energy all kinetic at bottom of swing
Roller coaster example
Consider the roller coaster in the figure. What
speed is needed at point A to reach point B?
Answer: E = KEA + PEA = KEB + PEB => ½mv2
+ mg(15 m) = mg(25 m) => ½mv2 = mg(10 m)
=> v2 = 2g(10 m) => v= 14.0 m/s
Your turn: Find the speed at point C.
Answer: E = KEB + PEB = KEC + PEC =>
mg(25 m) = ½mv2 + mg(0 m) =>
½mv2 = mg(25 m) => v2 = 2g(25 m) =>
v= 22.1 m/s
Conservation of energy
• If only conservative forces do work, then
the total mechanical energy of the system
remains constant.