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Phys 2101
Gabriela González
Work-energy theorem: The change in kinetic energy
of a particle is equal to the net work done on the
particle by all external forces.
 
1 2 1 2
W = ∫ F • dx = ΔKE = mv a − mv b
2
2
a
b
Consider a constant 1-D force producing constant acceleration a, starting from
rest. In time t, the work done by the force is Fd, where d is the distance
traveled, ½ a t2. At time t, the velocity is v=at, so the kinetic energy is
€
K = ½ mv2 = ½ m(at)2 = (ma) (½ a t2) = Fd
2
A variable force F acts along the x-axis on a 10kg mass. The particle
starts moving to the right with speed 2 m/s at x=0.
a)  Describe the motion of the mass.
b)  At what position does the particle has maximum and minimum kinetic
energy?
c)  Plot the position versus time for the particle.
d)  What is the expression for the force as a function of position?
F(N)
20
10
-2 -1
0
-10
1
2
x(m)
-20
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Hooke’s law:


F = −k d
Key concepts:
•  d: displacement from free, relaxed end
•  -ve sign: “restoring” force
•  large k: “stiff”, small k= “soft”
In one dimension, with the origin at the free,
relaxed end:
F=-kx
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•  Gravitational force: magnitude = mg, always vertical, pointing down.
•  sometimes does positive work, sometimes negative work.
•  Normal force: magnitude =whatever is needed for the objects not to
penetrate surface; always perpendicular to surface.
•  never does any work !!
•  Friction force: magnitude = µκN (if moving), ≤ µsN (if not); always
opposing motion and parallel to the surface.
•  always negative work ! (or no work, if there is no motion).
•  Tension: magnitude determined by acceleration; always parallel to
string or cord.
•  does positive work when lifting or pulling, negative work when
lowering or resisting pull.
•  Spring forces: magnitude = - kx; always parallel to spring, trying to
“restore” its relaxed state.
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•  sometimes does positive work, sometimes negative work.

xf
xf

xi
xi
 
W = ∫ F • dx = ∫ F dx
•  If force is constant (gravity, tension, friction),
xf
W = ∫ F dx = F (x f − xi )
xi
•  For a spring force,
xf
xf
1 2
W = ∫ F dx = ∫ − k x dx = − k x f − xi2
2
xi
xi
(
)
Work-energy theorem valid for all forces:
The change in kinetic energy of a particle is equal to the net work
done on the particle by all external forces.
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•  Power = energy delivered by a force F per unit time
•  Average power: P=ΔW/Δt
•  Instantaneous power: P=dW/dt
•  Constant force : W=Fx, P=dW/dt = F dx/dt = Fv
•  Spring force F=-kx, W= - ½ k x2
P = dW/dt = - ½ k (2 x) dx/dt = - k x v = Fv
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•  A force is conservative if the work it does on a
particle when coming back to the initial point is
zero.
Any constant force (like gravity) is a conservative
force (why?), friction is not (why?).
•  When a conservative force does work W on an
object, it changes the amount of potential energy of
that object by
ΔU = − W
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ΔU = − W
•  If the force does negative work, it increases the potential
energy.
•  If the force is constant,
W = F·d and ΔU = -F·d.
•  If the force is the (constant) gravitational force during vertical
motion, and if the y-axis points up, then
ΔU = -(-mg) (yf-yi) = +mgΔy
if particle goes up, gravity does negative work
(Wg=-mgΔy<0), the particle’s potential energy increases
(ΔUg=+mgΔy>0).
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€
When the projectile hits the ground…
•  Is the work done by gravity positive or negative?
•  Is the change in potential energy positive or negative?
•  Is the kinetic energy the same, larger or smaller than initially?
•  What is the projectile’s speed?

xf

  
W g = ∫ Fg • dx = F • d = mgh

xi
y
ΔU g = −Wg = −mgh
ΔKE = Wg
1
2
mv 2f − 12 mv02 = mgh
⇒ v 2f = v02 + 2 gh
mg
v
d
x
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The spring does work on a mass attached to its end:
there is energy transferred from the spring to the
mass, and converted into kinetic energy.
v
spring does negative work,
potential energy increases,
kinetic energy decreases,
mass is slowing down
spring does positive work,
potential energy decreases,
kinetic energy increases,
mass is speeding up
HRW simulation
F
v
F
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Potential energy in the mass due to the work of
spring, when the mass is at position x :
x
x
(
ΔU = −Ws = − ∫ Fs dx = ∫ kx dx = 12 k x 2 − x02
x0
)
x0
If we choose as initial position the origin, where the
spring is relaxed, then U=½ k x2.
Work energy theorem:
Conservation of
energy!
ΔKE = Ws
1
2
m(v
1
2
mv 2 + 12 kx 2 = 12 mv02 + 12 kx02
2
− v )= − k (x
2
0
1
2
2
2
0
−x
)
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•  There is potential energy associated with the work of conservative
forces (gravity: U=mgh; springs: U= ½ kx2):
ΔU=Uf-Ui=-W
•  If the only forces doing work are conservative the sum of kinetic and
potential energy is conserved:
ΔU+ΔK = 0
Uf+Kf = Ui+K
•  If there are non-conservative forces, the change in the total energy
(potential + kinetic) is equal to the work done by those forces:
ΔU+ΔK = Wother
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If only conservative forces are in action (gravity, springs,
normal forces, tension…) then the mechanical energy,
the sum of potential and kinetic energy, is conserved.
maximum kinetic = minimum potential
minimum kinetic = maximum potential
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A small block of mass m is released from rest at point P, at
a height h=5R above the bottom of the frictionless loop.
(a)  How much work does gravity do on the block from P to
Q?
(b)  How much to the top of the loop?
(c)  What is the net force acting on the
block at point Q?
(d)  What should be h if the block at the
top of the loop is on the verge of
losing contact?
(e)  What happens if there is an initial
velocity?
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