Transcript work, spring PE

Conservation of Energy
The work-energy theorem says
W  U  K  E
If no external force does any work on an object
(or system), then the energy of the system does
not change
The total energy is the same at beginning and
end (and all times between)
The energy of an isolated system is
conserved
Work
Work is the transfer of energy to an object
by applying external force.
Work is not a vector. It has a sign, but
no direction in space.
The work done by a force, on an object,
is positive if the force is giving energy to
the object.
depends on the size of the force and the
displacement through which it is applied.
In 1-D, if force is constant:
W  Fext d
Work — Units
Units of work or of energy
Since W  F d
ext
Units of work = (units of force)*(unit of
distance)
Units of work = N m
also called a Joule
1 J 1 N m
Work—Lift Example
W  Fext d
I lift a 2-kg weight up onto a 1.1 m-tall table.
How much work do I do?
a) Suppose I lift straight up at constant
speed.
b) Suppose slide up a 22o slope at constant
speed.
Lift Example 1
a) Given d = 1.1m, what F do I need to exert?
Constant velocity  No net force, so my
force to lift each is
F = mg = (1kg)(9.8m/s2)
Work done lifting each:
W  Fext d  (9.8 N)(1.1 m)  10.8 J
Note: F and d in same direction (both positive
or both negative) make W positive.
Total work done lifting both: 21.6 J
Lift Example 2
b) Given h = 1.1m, d=1.1/sin(22o) = 2.936 m
what F do I need to exert? Sum of forces in xdirection (along slope) =0, so
F = mgsin(q) = (2kg)(9.8m/s2)sin(22o) =7.342N
Work done sliding:
W  Fext d  (7.342 N)(2.936 m)  21.6 J
Total work done sliding: 21.6 J
Lift Example 3
I can’t lift the weight at constant velocity, if it
isn’t moving to begin with. Suppose I
accelerate it at a rate of 2.1m/s2 for 0.25 s,
then lift at constant speed, then slow it
back down at a rate of 2.1 m/s2 for 0.25 s.
No extra work is done overall in speeding it up,
then slowing it back down to its original
speed (here zero).
Lift Example 3
Part 1- speeding it up. SF=ma F-mg=ma, so
F= mg+ma = (2kg)(9.8m/s2)+(2kg)(2.1m/s2) = 23.8N
d=½ at2 =½ (2.1m/s2) (.25)2 = 0.065625m
W1=(23.8N)(0.065625m)= 1.561875J
Part 3 – slowing down SF=ma F-mg=ma, so
F= mg+ma = (2kg)(9.8m/s2)+(2kg)(-2.1m/s2) = 15.4N
d again= 0.065625m
W3=(15.4N)(0.065625m)= 1.010625 J
Part 2 – constant speed SF=ma F-mg=0, so
F= mg = (2kg)(9.8m/s2) = 19.6N
d = rest of the distance = 1.1m – 2(0.065625m) = 0.96875 m
W2=(19.6N)(0.96875 m)= 18.9875 J
Total work = 1.561875J+ 1.010625 J+ 18.9875 J = 21.6J!
Coincidence? No!
Total work done raising weight is 21.6 J,
whether you lift straight up or up a slope, at
constant speed or not, …
because the weights have gained 21.6 J of
energy from me, independent of how I go
about giving it to them.
21.6 J is how much energy it takes to make
this particular change (raising) in the state
of the weights, no matter what.
Work on a Spring
W = Fd
Only works if F is constant throughout
displacement,
Spring force is not constant
F=-kx
Work on a Spring
Calculus to the rescue: If we consider a small
displacement, dx, over which the force is
_________________, then the small amount of
work, dW, is
dW = Fdx
To get the total work,
xf
W   kxdx  kx  kx
1
2
xi
2
f
1
2
2
i
Spring Potential Energy
Once again, if the force does work on the spring

Where does that energy go?
Like with gravity, it is
W  U spring  kx  kx
1
2
2
f
1
2
2
i
Work in 2-D and 3-D
W  F d
Two ways to calculate this:
W=Fxdx+Fydy+Fzdz
using components of the vectors,
or W=Fdcos(f)
where F and d are magnitudes, and f =
angle BETWEEN F and d
NOT angle from x-axis
Conservation Example
A 35-g ball is placed on a compressed spring,
which shoots it straight up. The spring has a
spring constant of 220 N/m, and is initially
compressed by 3.5 cm. Neglecting drag, how
high does the ball go?
Do we have an isolated system?
Use: U
i
 Ki  U f  K f
h = 0.393 m or 39 cm above where it started
Drag
 Drag is resistance to motion of an object
through a fluid
If fluid is air, sometimes called air resistance
Drag with streamline, non-viscous flow
depends on:
fluid density (r), cross-sectional area of
object (A), speed of object relative to fluid
(v), properties of object’s surface (C).
Cross-sectional area can be thought of as
the area of the fshadow
the object would
 N
have, if lit from the direction of the passing
fluid.
s , max
s
Drag
 Depends on:
density of fluid (r), cross-sectional area of
object presented to fluid (A), relative
speed of object and fluid (v), properties of
the object’s surface (C).
D  C r Av
1
2
2
 Direction: Always opposes relative motion of
fluid and objectf   N
s , max
s
Note: This eqn doesn’t apply to viscous or
turbulent flow
Projectiles and drag
An object moving vertically does have the
same vertical motion as an object that is
moving sideways too, even if vyi is same, if
drag is not negligible.
Drag force has a vertical component that
depends on speed, not just vy
f s ,max  s N