Transcript (½)m(v 2 )

Sect. 8-4: Solving Problems with Energy Conservation
Conservation of Mechanical Energy
 K + U = 0 (Conservative forces ONLY!!)
or E = K + U = Constant
• For gravitational U:
Ugrav = mgy
E = K1 + U1 = K2+ U2
 (½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2
y1 = Initial height, v1 = Initial velocity
y2 = Final height, v2 = Final velocity
Conservation of Mechanical Energy
 K + U = 0 (Conservative forces ONLY!!)
or E = K + U = Constant
• For elastic (Spring) U:
Uelastic = (½)kx2
K1 + U1 = K2+ U2
 (½)m(v1)2 + (½)k(x1)2 = (½)m(v2)2 +(½)k(x2)2
x1 = Initial compressed (or stretched) length
x2 = Final compressed (or stretched) length
v1 = Initial velocity, v2 = Final velocity
Example 8-6: Pole Vault
Estimate
the kinetic energy & the speed
needed for a 70 kg pole vaulter to
just pass over a bar 5.0 m high.
Assume that the vaulter’s center of
mass is initially 0.90 m off the ground
& that he reaches his maximum height
at the level of the bar.
Example 8-7: Toy Dart Gun
For elastic forces, Mechanical energy Conservation gives:
A dart of mass m = 0.1 kg is pressed
against the spring of a toy dart gun.
The spring (spring constant k = 250
N/m) is compressed x = 6.0 cm &
released. The dart detaches from the
spring when the spring reaches its
natural length (x = 0). Calculate the
speed of the dart there.
x1 = 0.06 m, v1 = 0, m = 0.1 kg
k = 250 N/m, x2 = 0, v2 = ?
Get: v2 = 3 m/s
Example 8-8: 2 Kinds of Potential Energy
A ball, of mass m = 2.6 kg, s
s starts from rest, falls a distance
e h = 55.0 cm before striking a
a vertical coiled spring, which it I
I compresses an amount
l Y = 15.0 cm.
C
v1 = 0
v2 = ?
a Calculate the spring constant k.
a The spring has negligible mass.
a Ignore air resistance. Measure all a
a distances from the point where the a
a ball first touches the uncompressed a
a spring (take y = 0 at that point).
v3 = 0
v1 = 0
m = 2.6 kg, h = 0.55 m
Y = 0.15 m, k = ?
A two step problem:
STEP 1: (a)  (b)
v2 = ?
 (½)m(v1)2 + mgy1 =
(½)m(v2)2 + mgy2
v1 = 0, y1 = h = 0.55 m, y2 = 0
Get: v2 = 3.28 m/s
STEP 2:
(b)  (c)
(both gravity & spring U)
v3 = 0
 (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)
Example: Bungee Jump
m = 75 kg, k = 50 N/m, y2 = 0
v1 = 0, v2 = 0, y1 = h = 15m + y
y = ?
Conservation of Mechanical Energy
with both gravity & spring (elastic) U
 (½)m(v1)2 + mgy1
= (½)m(v2)2 + mgy2 +(½)k(y)2
= 0 + mg(15 + y) = 0 + 0 + (½)k(y)2
A quadratic Equation for y:
Solve & get y = 40 m & -11 m
(throw away the negative value)
Solve by Fig. (a)  to Fig. (c) directly!!



Δy + 15m




Example 8-9: A swinging pendulum
The simple pendulum consists of a small bob of mass m suspended
by a massless cord of length ℓ. The bob is released (without a push) at
t = 0, where the cord makes an angle θ = θ0 to the vertical.
a. Describe the motion of the bob in
terms of kinetic energy & potential
energy.
b. Calculate the speed of the bob as a
function of position θ as it swings back
& forth.
c. Calculate the speed at the lowest point
of the swing.
d. Calculate the tension in the cord, FT.
Example – A Spring Loaded Cork Gun
•
•
•
A ball of mass m = 35 g = 0.035 kg in a popgun is
shot straight up with a spring of constant k. The
spring is compressed yA = - 0.12 m, below it’s
relaxed level, yB = 0. The ball gets to maximum
height yC = 20.0 m above the relaxed end of the
spring.
(A) If there is no friction, find the spring constant k.
(B) Find speed of ball at point B.
It starts from rest & speeds up as the spring pushes
against it. As it leaves the gun, gravity slows it down.
Only conservative forces are acting, so we use
Conservation of Mechanical Energy
The initial kinetic energy K = 0. Choose both the
gravitational potential energy Ug = 0 & elastic
potential energy Ue= 0 where the ball leaves the gun.
At it’s maximum height, K = 0 again.
Note that
Two types of potential energy are needed!
• For the ball’s entire trip,
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 the 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 the 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
Other forms of energy; Energy Conservation
Nonconservative, or dissipative, forces:
Friction, Heat, Electrical energy, Chemical energy &
more do not conserve mechanical energy. However,
when these forces are taken into account, the total
energy is still conserved:
Sect. 8-5: The Law of Conservation of Energy
The Law of Conservation of Energy is one of the most
important principles in physics.
The total energy is neither decreased nor
increased in any process.
Energy can be transformed from one form to another
& from one body to another, but the
total amount remains constant.
 Law of Conservation of Energy
• Again: Not exactly the same as the Principle of Conservation of
Mechanical Energy, which holds for conservative forces only!
This is a general Law!!
Sect. 8-6: Problems with Friction
• We had, in general:
WNC = K + U
WNC = Work done by non-conservative forces
K = Change in Kinetic Energy
U = Change in Potential Energy (conservative forces only!)
• Friction is a non-conservative force! So, if
friction is present, we have (WNC  Wfr)
Wfr = Work done by friction
In moving through a distance d, the force of friction
Ffr does work Wfr = - Ffrd
When friction is present, we have:
Wfr = -Ffrd = K + U = K2 – K1 + U2 – U1
– Also now, K + U  Constant!
– Instead, K1 + U1+ Wfr = K2+ U2
OR:
K1 + U1 - Ffrd = K2+ U2
• For gravitational U:
(½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 + Ffrd
• For elastic or spring U:
(½)m(v1)2 + (½)k(x1)2 = (½)m(v2)2 + (½)k(x2)2 + Ffrd
Example 8-10: Roller Coaster with Friction
A roller-coaster car, mass
m = 1000 kg, reaches a vertical
height of only y = 25 m on the
second hill before coming to a
momentary stop. It travels a
total distance d = 400 m.
Calculate the work done by friction (the thermal energy produced) &
estimate the average friction force (assume it is roughly constant) on
the car.
m = 1000 kg, d = 400 m, y1 = 40 m, y2= 25 m, v1= y2 = 0, Ffr = ?
(½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 + Ffrd

Ffr= 370 N
Example 8-11: Friction with a Spring
A mass m slides along a
rough horizontal surface at a
speed v0. It strikes a massless
spring head-on & compresses
it a maximum distance X.
The spring constant is k.
Calculate the coefficient of
kinetic friction μk between
the block surface.
Example – Block Pulled on a 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 – A 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.
Example – A Crate Sliding Down a Ramp
• A crate, mass m = 3.0 kg, starts
from rest at height y0 = 0.5 m &
slides down a ramp of length
d = 1.0 m & incline angle
θ = 30° under a constant friction
force Fk = 5 N. It continues to
move on the horizontal surface
afterwards.
(A) Find the speed at the bottom.
(B) Assuming the same friction force,
find the distance on the horizontal
a surface that the crate moves after
a it leaves the ramp.
Example: A Block-Spring Collision
• Block, mass m = 0.8 kg, is given
an initial velocity vA = 1.2 m/s to
the right. It collides with spring
with constant k = 50 N/m.
(A) If there is no friction, find the
maximum compression distance
xmax of spring after the collision.
(B) Suppose there is a constant
friction force Fk between the
block & the surface. The
coefficient of friction is μk = 0.5.
Find the maximum compression
distance xC now.
Example: Connected Blocks in Motion
• Two blocks, masses m1 & m2,
are connected by spring of
constant k. m1 moves on a
horizontal surface with
friction. The system is released
from rest, when the spring is
relaxed and m2 is at height h
above the floor. The motion
eventually stops when m2 is on
floor. Calculate the kinetic
friction coefficient μk between
m1 & table.