Transcript f (x) - mrdsample

AP C UNIT 3
WORK & ENERGY
SCALAR PRODUCT
or DOT PRODUCT
Dot Product is defined
as the

magnitude of 1st ( a ) times
scalar
component of 2nd vector

( b ) along direction of 1st.
Essentially, the dot product
gives you information about how
much of each vector lies along
the direction of the other.
 
a  b  ab cos 
it’s a scalar result where Ф is
the angle between a and b.

a


b
THIS IS NOT THE SAME
AS ADDING 2 VECTORS
YIELDING A RESULTANT
The reason the dot product is
used in physics is because the
operation between certain vector
quantities produce meaningful
physical answers such as WORK.
Since the dot product involves the cosine then
IF Ф = 90o THEN PRODUCT = 0
IF Ф = 0o THEN PRODUCT = MAXIMUM
This is consistent in that when vectors are perpendicular neither lies along
the other, therefore an answer of zero results.
Directional properties of the Dot Product include:






i  i  j j  k  k  1

and





i  j  j k  i  k  0
Unit vector form of dot product:






 
a  b  (a x i  a y j  a z k ) (bx i  by j  bz k )
if you distribute, this would reduce to…
 
a  b  axbx  a y by  az bz
*Note that there is no direction associated with result but
answer can be negative depending on angle.
Calculate the dot product of the following
vectors and find the angle between them:
A = -3i + 5j
B = 6i +14j
More Calculus  - Derivative of a Product:
When taking the derivative of two functions multiplied
together, the derivative is: The 1st function times the derivative
of the 2nd plus the derivative of the 1st times the 2nd function.
d
dv
du
uv  u  v
dx
dx
dx

 u v  v u
'
'
Example…find dy/dx
y  (3x  1)( x  2)
2
u
v
y  (3x 2  1)( x  2)
dy
 (3x 2  1)  1  ( x  2)  6 x
dx
dy
2
 9 x  12 x  1
dx
*Could have ‘foiled’ and then performed
power rule as well
Chain Rule:
The chain rule is used when there is a function within a function.
f‘ (x) = f‘( g(x) ) (g'(x))
f (x) = f ( g(x) )
Think of the functions f and g as ``layers'' of a problem.
Function f is the ``outer layer'' and function g is the ``inner
layer.'' Thus, the chain rule tells us to first differentiate the outer
layer, leaving the inner layer unchanged (the term f'(g(x))) ,
then differentiate the inner layer (the term g'(x) ).
f ( x)  ( x  5)
3
The inner layer, g(x) is
3
1
2

f  2( x  5)  3x
2
(x3
+ 5)
The outer layer f(x) is (x)2
Derivative of outer times derivative
of inner with respect to x
f   6 x  30 x
5
2
Example:
Find the
derivative of f’
function of a
function
1
f ( x)  2
x 3
Work Done by a Constant Force
Work is defined as an external force
(F) moving through a displacement
(Δr). How much force lies along the
movement of an object.
Positive & Negative Work
In all 4 cases, the force has the same magnitude
and the displacement of the object is to the right
with the same magnitude. Rank the situations from
most positive to most negative.
A crate of mass, M, is dragged along a level
rough surface a distance, x, by a force, F as
shown. The coefficient of friction is uk.
Find the net work done on the crate in
terms of given variables and constants.
Work done by a varying force
x2
W   F ( x)dx
x1
If F(x) = 4x2 then find
the work done on a
particle that moves from
x = 1m to x = 5m.
Example:
Suppose a mass moves with a trajectory defined by
the position vector
r (t )  (te
t

10
2 ˆ
ˆ
)i  (t  3t ) j
Find the work done by the force, F  10iˆ  4 ˆj
over the interval from t = 1 to t = 2.
Work-Energy Theorem
x2
W   F ( x)dx
x1
A force, F(x) = 2-4x, acts on a 7.0kg mass. What is
the final speed of the mass as it is moved from x=5m
to x=2? Assume mass starts from rest at t=0.
Hooke’s Law
Work done by Spring
Negative means that the force opposes
the displacement from equilibrium
If block is pulled to right, the
force by spring is NOT
constant via Hooke’s Law.
Therefore, the work done by
spring must use avg force or
be integrated as:
Fs
A plot of spring force vs
displacement reveals a slope
equal to spring constant, k
k1
k1
kk 2
k1
2
k2
Two springs are attached to a block in series and
parallel as shown above. Determine the effective spring
constant for each situation in terms of k1 and k2 .
What would spring
constant be if a mass was
attached to a massless
spring that stretched a
distance x?
What if same mass was
attached to 2 springs as
shown? How would stretch,
x, differ for each spring?
Power
Rate at which work is done or energy is
transferred
A 4kg particle moves along the x-axis. Its position
varies with time according to x = t + 2t3, where x is in
meters and t is in seconds. Find the the power
being delivered to the particle at any time t
OR
If a projectile thrown directly upward reaches a
maximum height h and spends a total time in the air T, the
average power of the gravitational force during the
trajectory is:
a) 2mgh / T
b) -2mgh / T
c) 0
d) mgh / T
e) -mgh / T
Potential Energy
As height above Earth increases…
Conservative &
Non-conservative Forces
The work a conservative force does on an object in
moving it from A to B is path independent - it depends
only on the end points of the motion.
Force of gravity and the spring force are conservative
forces. Conservative forces ‘store’ energy…available
for kinetic energy
The work done by non-conservative (or dissipative)
forces in going from A to B depends on the path
taken. Friction is non-conservative. Nonconservative forces don’t ‘store’ energy.
CONSERVATION OF NRG
Total energy in a closed system remains unchanged
Work done by NC forces or friction is positive in the
above formula. No need to put in minus sign. Work
done by friction occurs on left side as minus but
becomes + when taken to other side.
Energy worksheet
Potential Energy Function
Diagrams
U(x) vs Restoring Force
Relationship
As mass oscillates
the net force is
composed of
gravity and the
spring force. The
net restoring force
is shown in the
graph.
Net force at
equilibrium = 0.
When slope of
U(x) is negative,
force is positive
and vice versa.
Potential Energy and Conservative
Restoring Forces (gravity, spring)
The instantaneous
restoring force is equal to
the negative derivative of
the potential energy
function.
In the case of a mass oscillating on a
horizontal frictionless surface we can
verify this relationship:
Example1
A certain spring is found not to obey Hooke’s Law; it
exerts a restoring force F(x) = -60x-18x2.
a) Calculate the potential energy function U(x) for
this spring. Let U = 0 when x = 0.
b) An object with mass 0.90kg on a frictionless,
horizontal surface is attached to this spring, pulled a
distance 1.00m to the right to stretch the spring and
released. What is the speed of the object when it is
0.50m to the right of x=0?
Potential energy
diagram states of
equilibrium:
Points of equilibrium
are where the force is
zero (slope = zero).
x3 and x5 are points of stable equilibrium or energy wells. If the system is
slightly displaced to either side the forces on either side will return the
object back to these positions.
x6 is a position of neutral equilibrium. Since there is no net force acting
on the object (slope of U(x) = 0) it must either possess only potential
energy and be at rest or, it also possesses kinetic energy and must be
moving at a constant velocity.
x4 is a position of unstable equilibrium. If the object is displaced ever so
slightly from this position, the internal forces on either side will act to
encourage further displacement instead of returning it back to x4.
Example2
A 5-kg mass moving along the xaxis passes through the origin with
an initial velocity of 3m/s. Its
potential energy as a function of its
position is given in the graph.
a) How much total energy does
the mass have as it passes
through the origin?
b) Between 2.5m and 5m, is the
mass gaining speed or losing speed?
c) How fast is it moving at 7.5m?
d) How much potential energy would have
to be present for the mass to stop moving?
Turning points
Positions where
potential energy
equals the total
mechanical energy,
Umax= E, are called
turning points
A particle moves along the x-axis according to
the following potential energy function:
2.4 Nm
U ( x)  (0.60 N ) x 
x
2
Find the positions of equilibrium for the particle.
Find F(x) when
 ax
U ( x)  2
2
b x