Download Chapter 8 Examples

Chapter 8
Conservation of Energy
EXAMPLES
Example 8.1 Free Fall
(Example 8.1 Text book)
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Determine the speed of the ball at y above
the ground
The sum K + U remains constant
At h: Ui = mgh Ki = 0
At y: Kf = ½mvf2 Uf = mgy
In general: Conservation of Energy
Ki + Ui = Kf + Uf 
0 + mgh = ½mvf2 + mgy 
Solving for vf v 2  2 g h  y   v  2 g h  y 
f
f
vf is independent of the mass !!!
Example 8.2 Roller Coaster Speed
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Using Mechanical Energy
Conservation (Frictionless!)
½mvf2 + mgyf = ½mvi2 + mgyi
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Only vertical differences matter!
Horizontal distance doesn’t matter!
Mass cancels!
Find Speed at bottom?
Known: yi = y = 40m, vi = 0,
yf = 0, vf = ? 
0 + mgyi = ½mvf2 + 0 
vf2 = 2gyi = 784m2/s2 
v2 = 28 m/s
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Example 8.3 Spring-Loaded Gun
(Example 8.3 Text book)
Choose point A as the initial point and C as
the final point
 (A). Find the Spring Constant k ?
Known: vA = 0, yA = 0 , xA = yB = 0.120m
vC = 0 , yC = 20m, UC = mgyC , m = 35.0g
EC = EA 
KC + UgC + UsC = KA + UgA + UsA
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½mvC2 + mgyC + ½kxC2 = ½mvA2 + mgyA + ½kxA2
 0 + mgyC + 0 = 0 + 0 + ½kxA2
 ½kxA2 = mgyC  k = 2mgyc/xA2
=2(0.0350kg)(9.80m/s2)(20.0m)/(0.120m)2
k = 953 N/m
yA
yC
yB
Example 8.3 Spring-Loaded Gun, final
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(B). Find vB ?
yC
Use: EB = EA
KB + UgB + USB = KA + UgA + USA
½mvB2 + mgyB + ½kxB2 =
½mvA2 + mgyA + ½kxA2 
½mvB2 + mgyB + 0 = 0 + 0 + ½kxA2
vB2 = (kxA2 – 2mgyB)/m
vB2 =388.1m2/s2 
vB = 19.7 m/s
yB
yA
Example 8.4 Ramp with Friction
(Example 8.7 Text book)
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Problem: the 3.0 kg crate slides down the rough
ramp. If: vi = 0, yi = 0.5m , yf = 0
ƒk = 5N
 (A). Find speed at bottom vf
At the top: Ei = Ki + Ugi = 0 + mgyi
At the bottom: Ef = Kf + Ugf = ½ m vf2 + 0
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Recall: If friction acts within an isolated system
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ΔEmech = ΔK + ΔU = Ef – Ei = – ƒk d 
½ m vf2 – mgyi = – ƒk d
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Solve for vf
v 2f 
2
mgyi  f k d   6.47m 2 / s 2  v f  2.54m / s
m
Example 8.4 Ramp with Friction, final
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(B). How far does the crate slide on the horizontal
floor if it continues to experience the same friction
force ƒk = 5N
The total ΔEmech is only kinetic (the potential energy
of the system remains fixed):
ΔEmech = ΔK = Kf – Ki = – ƒk d
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Where:
Then:
Kf = ½ m vf2 = 0
Ki = ½ m vi2 = 9.68 J
Kf – Ki = 0 – 9.68 J = – (5N)d 
d = (9.68/5) = 1.94 m
Example 8.5 Motion on a Curve Track
(Frictionless)
A child of mass m = 20 kg starts sliding
from rest. Frictionless!
 Find speed v at the bottom.
 ΔEmech = ΔK + ΔU
ΔEmech =(Kf – Ki) + (Uf – Ui) = 0
(½mvf2 – 0) + (0 – mgh)= 0
½mvf2 – mgh = 0 
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v f  2 gh  2(10)(2)  6.3m / s
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Same result as the child is falling
vertically trough a distance h!
Example 8.5 Motion on a Curve Track
(Friction)
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If a kinetic friction of ƒk = 2N acts on the child and the
length of the curve track is 50 m, find speed v at the
bottom.
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If friction acts within an isolated system
ΔEmech = ΔK + ΔU = – ƒk d
ΔEmech = (Kf – Ki) + (Uf – Ui) = – ƒk d
ΔEmech = ½mvf2 – mgh = – ƒk d
ΔEmech = ½(20) vf2 – 20(10)(2) = –100J
2
v  mgyi  f k d   30m 2 / s 2  v f  5.5m / s
m
2
f
Example 8.6 Spring-Mass Collision
(Example 8.8 Text book)
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Frictionless!
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K +Us = Emech remains constant
(A). Assuming: m= 0.80kg vA = 1.2m/s
k = 50N/m
Find maximum compression of the spring
after collision (xmax)
EC = EA  KC + UsC = KA + UsA
½mvC2 + ½kxmax2 = ½mvA2 + ½kxA2
0 + ½kxmax2 = ½mvA2 + 0 
xmax 
m
0.80kg
vA 
(1.2m / s)  0.15m
k
50 N / m
Example 8.6 Spring-Mass Collision, final
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(B). If friction is present, the energy decreases by:
ΔEmech = –ƒkd
Assuming: k= 0.50 m= 0.80kg vA = 1.2m/s k = 50N/m
Find maximum compression of the spring after collision xC
ΔEmech = –ƒk xC = –knxC = –kmgxC 
ΔEmech = –3.92xC
(1)
Using: ΔEmech = Ef – Ei
ΔEmech = (Kf – Uf) + (Ki – Ui)
ΔEmech = 0 – ½kxC2 + ½mvA2 + 0
ΔEmech = – 25xC2 + 0.576 (2)
Taking: (1) = (2): – 25xC2 + 0.576 = –3.92xC
Solving the quadratic equation for xC :
xC = 0.092m < 0.15m (frictionless)
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Expected! Since friction retards the motion of the system
xC = – 0.25m Does not apply since the mass must be to the right of the
origin.
Example 8.7 Connected Blocks in Motion
(Example 8.9 Text book)
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The system consists of the two blocks, the spring,
and Earth. Gravitational and potential energies are
involved
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System is released from rest when spring is
unstretched.
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Mass m2 falls a distance h before coming to rest.
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The kinetic energy is zero if our initial and final
configurations are at rest K = 0
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Find k
Example 8.7 Connected Blocks in Motion,
final
Block 2 undergoes a change in gravitational potential
energy
 The spring undergoes a change in elastic potential
energy
Emech = K + Ug + US = Ug + US = Ugf – Ugf + Usf –
Usi
Emech = 0 – m2gh + ½kh2 – 0
Emech = – m2gh + ½kh2 (1)
 If friction is present, the energy decreases by:
ΔEmech = –ƒkh = – km1gh (2)
 Taking (1) = (2):
– m2gh + ½kh2 = – k m1gh
k m1gh = m2gh – ½kh2 
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m2 g  12 kh
k 
m1 g