Transcript Part II

Examples: Mechanical Energy Conservation
Example 8.3 – Spring Loaded Cork Gun
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Ball, mass m = 35 g = 0.035 kg in popgun is shot
straight up with spring of unknown constant k.
Spring is compressed yA = - 0.12 m, below
relaxed level, yB = 0. Ball gets to a max height
yC = 20.0 m above relaxed end of spring.
(A) If no friction, find spring constant k.
(B) Find speed of ball at point B.
Ball starts from rest. Speeds up as spring pushes
against it. As it leaves gun, gravity slows it down.
System = ball, gun, Earth.
Conservative forces are acting, so use
Conservation of Mechanical Energy
Initial kinetic energy K = 0. Choose
gravitational potential energy Ug = 0
where ball leaves gun. Also elastic potential
energy Ue= 0 there. At max height, again
have K = 0. Choose Note:
Need two types of potential energy!
• For entire trip of ball,
Mechanical Energy is Conserved!!
or: KA + UA = KB + UB = KC + UC.
At each point, U = Ug + Ue so,
KA+ UgA+ UeA = KB+ UgB + UeB = KC + UgC + UeC
(A) To find spring constant k, use:
KA+ UgA+ UeA = KC + UgC + UeC
or, 0 + mgyA + (½)k(yA)2 = 0 + mgyC + 0, giving
k = [2mg(yC – yA)/(yA)2] = 958 N/m
(B) To find ball’s speed at point B, use:
KA+ UgA+ UeA = KB + UgB + UeB
or, (½)m(vB)2 + 0 + 0 = 0 + mgyA + (½)k(yA)2 , giving
(vB)2 = [k(yA)2/m] + 2gyA; or, (vB) = 19.8 m/s
Example: Toy Dart Gun
• Mechanical energy conservation!
 (½)m(v1)2
+ (½)k(x1)2
= (½)m(v2)2
+ (½)k(x2)2
• Speed when
leaves gun?
x1 = 0.06 m, v1 = 0, m = 0.1 kg, k = 250 N/m
x2 = 0, v2 = ? Find: v2 = 3 m/s
?
?
Example: Two Kinds of PE
m =2.6 kg
h =0.55 m
Y = 0.15 m
k=?
v1 = 0
v2 = ?
v3 = 0
A two step problem: STEP 1: (a)  (b)
 (½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2
v1 = 0, y1 = h = 0.55 m, y2 =0 . Find: v2 = 3.28 m/s
STEP 2: (b)  (c) (both gravity & spring PE)
 (½)m(v2)2+(½)k(y2)2+mgy2 = (½)m(v3)2 + (½)k(y3)2 + mgy3
y3 = Y = 0.15m, y2 = 0  (½)m(v2)2 = (½)kY2 - mgY
Solve for k & get k = 1590 N/m
ALTERNATE SOLUTION: (a)  (c) skipping (b)
Section 8.3: Problems with Friction
• We had, in general:
WNC = K + U
WNC = Work done by non-conservative forces
K = Change in KE
U = Change in PE (conservative forces)
• Friction is a non-conservative force! So, if
friction is present, we have (WNC  Wf)
Wf = Work done by friction
In moving through a distance d, force of kinetics
friction fk does work Wf = - fkd
When friction is present, we have:
Wf = -fkd = K + U = Kf – Ki + Uf – Ui
– Also now, K + U  Constant!
– Instead, Ki + Ui+ Wf = Kf + Uf
OR:
Ki + Ui - fkd = Kf+ Uf
• For gravitational PE:
(½)m(vi)2 + mgyi = (½)m(vf)2 + mgyf + fkd
• For elastic or spring PE:
(½)m(vi)2 + (½)k(xi)2 = (½)m(vf)2 + (½)k(xf)2 + fkd
Example: Roller Coaster with Friction
m=1000 kg, d=400 m, y1=40 m, y2= 25 m, v1= y2 = 0, fk = ?
(½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 + fkd

fk= 370 N
Ex. 8.4 – Block Pulled on Rough Surface
• A block, mass m = 6 kg,
is pulled by constant
horizontal force F = 12 N.
over a rough horizontal
surface. Kinetic friction
coefficient μk = 0.15.
Moves a distance Δx = 3
m. Find the final speed.
Example 8.6 – Block – Spring System
• A mass m = 1.6 kg, is
attached to ideal spring of
constant k = 1,000 N/m.
Spring is compressed
x = - 2.0 cm = - 2  10-2 m
& is released from rest.
(A) Find the speed at x = 0
if there is no friction.
(B) Find the speed at x = 0
if there is a constant friction
force fk = 4 N.
Ex. 8.7 – Crate Sliding Down a Ramp
• Crate, mass m = 3.0 kg, starts
from rest at height yi = 0.5 m
& slides down a ramp, length
d = 1.0 m & incline angle
θ = 30°. Constant friction force
fk = 5 N. Continues to move on
horizontal surface after.
(A) Find the speed at the bottom.
(B) Assuming the same friction
force, find the distance on then
horizontal surface that the crate
moves after it leaves the ramp.
Ex. 8.8 – Block-Spring Collision
• Block, mass m = 0.8 kg, gets
initial velocity vA = 1.2 m/s to
right. Collides with spring with
constant k = 50 N/m.
(A) If no friction, find the
maximum compression distance
xmax of spring after collision.
(B) There is a constant friction
force fk between block &
surface. Coefficient of friction is
μk = 0.5. Find the maximum
compression distance xC now.
Ex. 8.9 – Connected Blocks in Motion
• Two blocks, masses m1 &
m2, are connected by
spring of constant k. m1
moves on horizontal
surface with friction.
Released from rest when
spring is relaxed when m2
at height h above floor.
Eventually stops when m2
is on floor. Calculate the
kinetic friction coefficient
μk between m1 & table.