Download Work and kinetic energy

Work and kinetic energy
•  LA info session today at 5pm in UMC235
•  CAPA homework due tomorrow night.
1
Work
I apply a force of 2N in the x
direction to an object that
moves 5m in x. How much
work have I done on the object?
Fx (N)
2
1
−1
1
2
3
4
5
x (m)
! ! ! !
W = F ⋅ Δr = F Δr cosθ = F(Δr)cosθ
Since the displacement and force are in the same
direction, simply need to multiple them together.
So W = Fx・Δx = (2N)(5m) = 10 J
This is the area underneath the force vs position curve.
2
Work by varying force
Suppose we have a force
that varies in x as shown.
How do we determine the
work done (for example in
going from x=0 to x=5)?
Fx (N)
2
1
−1
1
2
3
4
5
x (m)
Can divide up path into a bunch of small pieces,
calculate the work done in each piece, and add them up.
If we take the limit as the size of the piece goes to 0,
we get an integral. It is also the area under the curve.
x2
W=
∫ F dx
x
x1
= area under force vs position curve
3
Work by spring force
Remember the force by a spring is F = −kx. Therefore, the
force on a spring when you compress or stretch it is F = kx.
What is the work done on a spring when stretching from
position x1 to x2?
W=
∫
x2
x1
Fx dx = ∫
x2
x1
1 2
kx dx = kx
2
x2
x1
1 # 2
= k $ x2 − x12 %&
2
If we start from the equilibrium
point (x1=0) then W = 12 kx 2
This is the area underneath the curve
(a right triangle with sides x and kx).
Fx (N)
kx
x (m)
x
4
Even more details about work
When working in more than one dimension, the work
is further
! !
r
x
modified from W = ∫ Fx dx to the line integral W = ∫ ! F ⋅ dr!
2
x1
2
r1
So at each small displacement dr multiply the displacement
by the force acting in that direction. Add up all of these
small works to get the total work.
r
For a car going around a corner, the force
causing the radial acceleration is always
perpendicular to the velocity, and therefore it is
!
always perpendicular to dr and will do no work.
Since velocity is usually perpendicular to
the normal force, most of the time a normal
force will not do work either.
5
Clicker question 1
Set frequency to BA
A block moves along the x-axis and
Fx (N)
is acted upon by a variable force as
2
shown in the graph. Between
1
x=0m and x=5 m, how much work
is done on the block?
1 2 3
−
1
A.  -1 J
−2
B.  0 J
C.  3 J
The work done by the force can be
D.  4 J
found from the area under the curve.
E.  5 J
4
5
x (m)
W = 2N ⋅ 2m − 12 2N ⋅1m = 4 J −1 J = 3 J
6
Clicker question 2
Set frequency to BA
A 1 kg mass is moved part way around
up
a square loop as shown. The square 1.0 m
0.5 m
is 1 m on a side and the final position
start
of the mass is 0.5 m below its original
position. Assume g = 10 m/s2. What
finish
is the work done by the force of gravity
during this journey?
No work from the horizontal segments since
A.  –10 J
force and displacement are perpendicular
B.  –5 J
! !
2
C.  0 J Upward segment: Wg, up = Fg ⋅ Δr = − (1 kg) 10 m/s ( 0.5 m ) = −5 J
! !
D.  5 J
2
W
=
F
⋅
Δ
r
=
1
kg
10
m/s
( )
(1 m ) =10 J
E.  10 J Downward segment: g, down g
(
)
(
)
Wg, total = Wg, up + Wg, down = 5 J
7
Clicker question 2
Set frequency to BA
A 1 kg mass is moved part way around
up
a square loop as shown. The square 1.0 m
0.5 m
is 1 m on a side and the final position
start
of the mass is 0.5 m below its original
position. Assume g = 10 m/s2. What
finish
is the work done by the force of gravity
during this journey?
Can also think of the 0.5 m upward segment
A.  –10 J
being canceled out by half of the 1 m
B.  –5 J
downward segment leaving just a 0.5 m
C.  0 J
downward segment
D.  5 J
!
!
Total
work
E.  10 J
Wg, total = Fg ⋅ Δr = (1 kg) 10 m/s2 ( 0.5 m ) = 5 J
by gravity:
(
)
8
Clicker question 3
Set frequency to BA
A 1 kg mass is moved on a totally
1.0 m
up
crazy path that ends up in the same
place as in the previous problem. Is
start
the work done by gravity during this
0.5 m
journey the same or different
finish
compared to the previous path?
A.  Same
No work comes from any horizontal motion
B.  Different
C.  Impossible to tell
With the exception of the 0.5 m of difference between the start
and end, all other upward and downward motion cancels out
Therefore, the work done by gravity is the same!
9
Kinetic energy
Can consider
by an individual force
! !work
! done
!
W = F ⋅ Δr = F Δr cosθ = F(Δr)cosθ
If you add up the work done by all the forces (or compute
the work done by the net force) you get the net work: Wnet.
By Newton’s second law, Fnet = ma = m(dv/dt) so
dv
Wnet = ∫ Fnet dx = ∫ m dx =
dt
dx
∫ m dv dt =
∫ mv dv
Doing this integral, we get:
Wnet =
∫
v2
v1
1
2
mv dv = mv
2 v2
v1
1
2
2
2
1
2
2
1
= mv − mv
1 2
We define the kinetic energy of an object as K = 10mv
2
Work-energy theorem
The work-energy theorem states ΔK = Wnet
! !
2
2
1
1
ΔK = 2 mv f − 2 mvi
W = F ⋅ Δr
If the net work is positive, kinetic energy is added to the
object and if the net work is negative, kinetic energy is
removed from the object.
This is always true! But remember the net
work means the work done by all forces (or
equivalently, the work done by the net force).
11
Clicker question 4
Set frequency to BA
A hockey puck sliding on a frictionless ice rink at 1 m/s slides onto
a rug someone left on the ice. The puck comes to rest after
moving 1 m on the carpet. Which forces did work (either positive
or negative) while the puck was on the carpet?
A.  normal force
B.  gravity
C.  friction
D.  more than one of A-C
The normal force and gravity are both
E.  None of A-C
perpendicular to the displacement.
Therefore they do no work.
Kinetic friction force is always opposite the velocity vector so
it always does negative work (reducing the kinetic energy).
12