Download More en the Work-Energy Theorem Mechanical Energy Alternate

Mechanical Energy
More en the Work-Energy Theorem
In a mechanical "system", the quantity "Mechanical
Energy" is often used. It has the symbol Em.
A EK - EK2 - EK1
Wnet * A EK
W n e t =E K 2 - EK1
Em= E K + E g
Wnet = | Ad | | Fnet | cos9
Remember
Ee
If a mechanical "system" loses no energy to the
surroundings in the form of heat, light, sound etc over
time, the total mechanical energy remains constant or
is conserved.
This is called the[.conservationtA
mechanical
energy.
., , ir*!**—
"P'H
O*^
-r-^—^-^-
Therefore we can create a new work-energy formula...
I Ad] |F n e t |cose = E K 2 - EK1
;.-•.;
+
• i v v;,x'j.,.. •;••••••••<. :•>•.-. ..• v..-^.:
v
. ,'f,> :•:., '0 !,< .:• >;i'•;•••> ><? >•>;'•• - : , -•
•yfi:'t;*k-: J>X'-, '.-:, >:{U"' : '^- . • • •
Alternate Definition of Work
• .s1
^
Suppose an object is moving in a direction given by its
displacement as shown. Suppose the net force is acting t
as shown. What component of the net force does no work
on.the object?
is perpendicular to
net 90°
A d . It does no work on the
object. Depending on 6 ,
net I along A d can only
Ad
apeed up or slow down the
net I
object. Therefore, work done
We can write ...
..by the net force only changes
vv
the kinetic energy of the
net -"^K
object.
This is called the work-energy theorem.
net go|0
Example: A 1260 kg car is moving at 72.0 km/h [E].
Suddenly the brakes are applied and a constant kinetic
friction force of 5850 N [W] brings the car to a stop. Using
the work-energy theorem, find the braking distance.
Given:
m =1260 kg
= 0m/s
v1 = 72.0 km/h = 20.0 m/s
f = 5850N
9 = 180
Unknown: | Ad | = ?
Formula:
Sub:
Ad | Fnet cosO = E K 2 - EK1
Note f K
supplie
the net
force in
this
Ad | | Fnet | cose = E K2
ease.
2
| Ad | (5850) cosl 80° = 0 - (1260) (20.0) 12
- 5850 | Ad | = - 252000
I Ad I = 43.1 m
Example: A 72,0 N applied force compresses a vertical
spring 12,0 cm. A small 21.0 g marble is placed at rest on
the compressed.spring. The compressed spring is then
used to project the marble straight up into the air. If friction
is negligible, to what maximum height does the marble
rise?
' . .
•
Apply conservation of mechanical
inical
energy: Em (start)= Em (max height)
tti
:htt
0
K1
E
g1
E
e1 = EK2
+ 0 + KX^/2 =
KX2/2 = mgh
600(.12) 2 /2 = 0.021(10)h
hmax = 20.6 m
j j
_
E
g2> + ^e2
3h+ 0
X=
0.120m
Choose a zero reference
level for gravitational
potential energy. Why
can't it be ground?
hmax =?
K = 600 N/m
t*
Work done by an applied force
in a system with friction
Critical thinking:
Can you explain why this equation is valid to find the
work done by an applied force introduced to a system
where energy is lost to the surroundings due to
friction?
Fs = 72 N
I
Wapplied = AEm +
®r 0.021 kg
'^^^^^^Mi • • 1
ground
Example: A 72.0 N applied force compresses a vertical
spring 12.0 cm. A small 21.0 g marble is placed at rest on
the compressed spring. The compressed spring is then
used to project the marble straight up into the air. If friction
is negligible, to what maximum height does the marble
rise?
Find K first:
K=FS/X =
K = 72 70.120
K = 600 N/m
Systems With Friction
Most mechanical systems have kinetic friction. The
short-cut formula for the work done by the kinetic
friction force is W = - fk d
The negative sign represents a loss of heat energy
from the system to the surroundings
How can we modify the conservation of mechanical
energy formula to take friction into account?
m=
0.021 kg
0.120m
Choose a zero reference
level for gravitational
potential energy. Why
can't it be ground?
ground
•
(start) ~ ^ ^
=
•
m (later)