phy1012f energy

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PHY1012F

ENERGY

Gregor Leighgregor.leigh@uct.ac.za

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ENERGYLearning outcomes:

At the end of this chapter you should be able to…Define and use the concepts of kinetic and potential energy.Use various pictorial representations to interpret problems involving the transformation of energy from one kind to another. Apply the law of conservation of mechanical energy to solve mechanical problems in conservative systems, including (in conjunction with the law of conservation of momentum) problems involving elastic collisions.

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ENERGYEnergy…

is an extremely abstract concept and is difficult to define;is a number (a scalar) describing the state of a system of objects (for an isolated system this number remains constant, i.e. the energy of the system is conserved);appears in many different forms, each of which can be converted into another form of energy in one or other of the transformation processes which underlie all activity in the Universe;is all there is! (Even matter is just a manifestation of “coagulated” energy: E = mc2.)

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ENERGY

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sin s smg ds mv dv

sdv dsm ds dt

net ssF ma

y

KINETIC ENERGY and GRAVITATIONAL POTENTIAL ENERGY

Consider an object sliding over an arbitrarily-shaped frictionless surface during some brief interval…

n

w

v

…as it moves from an initial height, yi, to a final height, yf …

s

yi

yfdy

dy = sin dsds

Newton II:

mg sinmg cos

sdvm dt

sinmg ss

dvmv ds

mg dy

(chain rule)

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… mvs dvs = –mg dy y

KINETIC ENERGY and GRAVITATIONAL POTENTIAL ENERGY

s

yi

yff f

i i

v ys sv ymv dv mg dy

½ mvf2 – ½ mvi

2

½ mvf2 + mgyf = ½ mvi

2 + mgyi

½ mv2 = K Kinetic energy – energy due to a body’s motion.

mgy = Ug (Gravitational) potential energy – energy due to a body’s position.

i.e. Kf + Ugf = Ki + Ugi

= –(mgyf – mgyi)

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Units: [kg m2/s2 = joule, J]Kinetic energy can never be negative.

K = ½ mv2

Ug = mgy

KINETIC ENERGY and GRAVITATIONAL POTENTIAL ENERGY

Units: [kg m/s2 m = kg m2/s2 = joule, J]

For Ug calculations, the y-axis represents the vertical. (i.e. y-values are heights.)Only changes in potential energy are meaningful, since the height at which potential energy is zero is arbitrary. (U can be negative.) Irrespective of the chosen origin, however, Ug will always be computed as the same value.

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MECHANICAL ENERGYThe sum of the kinetic and potential energies of a system is called the mechanical energy of the system:

Emech = K + U

We have shown that (in an isolated system, where Fnet

= 0) a body’s mechanical energy remains constant as it moves under the influence of gravity: Kf + Ugf = Ki + Ugi

Hence: Emech = K + U = 0

Law of conservation of mechanical energy

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CONSERVATION OF MECHANICAL ENERGYAlthough the system is isolated, forces within it can transfer energy between kinetic and potential forms.Using a ball in free fall as an example, the interchange between its kinetic and potential energies (and the conservation of mechanical energy) can be illustrated using energy bar charts…

K U Emech = K + U

Emech = 0

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HOOKE’S LAW

The spring constant, k, is a measure of the “stiffness” of the spring.The negative sign indicates a restoring force.s = L – L0 is the displacement from equilibrium.(Fsp)s is the magnitude of the force acting on either side of the spring.Hooke’s law is not universal. It is applicable only within the limits of each particular spring.We shall work only with ideal (massless) springs.

sp sF k s

s > 0

sp 0s

F

sp 0s

F

s < 0

(unstretched)

sse0

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s mgx k

A toy train is joined to a 2.0 kg block by a spring with k = 50

N/m. The coefficient of static friction for the block is 0.60. The spring is at its equilibrium length when the train starts off at 5 cm/s.When does the block slip?

y

x

spFn

w

sf

(Fnet)x = Fsp – fs = 0

kx – s mg = 0

The block starts to slip when fs reaches its maximum: fs max = sn

t = 4.7 s

(Fnet)y = n – w = 0

20.60 2.0 kg 9.8 m/s50 N/m 0.235 m

xv t

0.235 m0.05 m/s t

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ELASTIC POTENTIAL ENERGYA spring exerts a force which varies with the extension (or compression) of the spring. The acceleration it causes is thus not constant.

sse0

We can simplify matters, however, by considering only the before-and-after situations…

km vi

vf

si

sf

net ssF maNewton II: sdvm dt

sdv dsm ds dtk s ss

dvmv ds(chain rule)

s sk s ds mv dv

spF

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i.e. Kf + Usp f = Ki + Usp i

ELASTIC POTENTIAL ENERGY… mvs dvs = –ks ds

f f

i i

v ss sv smv dv k s ds

½ mvf2 – ½ mvi

2 = –½ k(sf)2 + ½ k(si)2

½ mvf2 + ½ k(sf)2 = ½ mvi

2 + ½ k(si)2

½ k(s)2 = Usp Elastic potential energy – energy due to a body’s position.

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½ k(s)2 = Usp

ELASTIC POTENTIAL ENERGYUnits: [(N/m)(m2) = kg m2/s2 = joule, J]Since s is squared, elastic potential energy is positive irrespective of whether the spring is extended or compressed.

We have now shown that (in an isolated system, where Fnet = 0) a body’s mechanical energy remains constant as it moves under the influence of gravity and/or on an ideal spring in the absence of friction.

Hence: Emech = K + Ug + Usp = 0

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VERTICAL SPRINGS

K Ug Usp

y

0

ye

Besides kinetic energy, bodies moving at the end of vertical springs have both gravitational and elastic potential energies.E.g. An initially compressed spring

launches a sheep vertically upwards…

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ELASTIC COLLISIONSAn idealised collision in which mechanical energy is conserved is called an perfectly elastic collision.Problem-solving strategy for elastic collisions:1. Sketch the situation.2. If necessary, divide the problem into separate parts

according to the principles which will be used to solve it. (Use before-and-after drawings for conservation parts).

3. Establish a coordinate system to match the motion.4. Define symbols for initial and final masses and

velocities, and list known information, identify desired unknowns.

5. Apply the laws of conservation of momentum and energy.

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A 200 g steel ball on a 1 m-long string is pulled sideways until the string is at an angle of 45°, at which point it is released. At the very bottom of its swing it strikes a 500 g steel block resting on a frictionless table. To what angle does the ball rebound?

0 = 45°y

0

L = 1 m

mA = 200 g

mB = 500 g B

A

A B A BA

B

(y0)A = ?

(v0)A = 0 m/s

3 = ?

Part 1: Cons. of EPart 2: Cons. of P and E

Part 3: Cons. of E

(y1)A = 0 m

(v1x)A = ?(v1x)B = 0 m/s

(v2x)A = ?(v2x)B = ?

(y3)A = ?(v3)A = 0 m/s

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A 200 g steel ball on a 1 m-long string is pulled sideways until the string is at an angle of 45°, at which point it is released. At the very bottom of its swing it strikes a 500 g steel block resting on a frictionless table. To what angle does the ball rebound?

0 = 45°y

0

L = 1 m (y0)A = L – L cos0

½mA(v1)A2 + mAg(y1)A = ½mA(v0)A

2 + mAg(y0)A A

B

(y1)A = 0 m

(v1)A = ?(y0)A = ?

(v0)A = 0 m/s

y0

½mA(v1)A2 + 0 = 0 + mAg(y0)A

= 2.40 m/s 1 0A A2v g y

= L(1 – cos0) = 0.293 m

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A 200 g steel ball on a 1 m-long string is pulled sideways until the string is at an angle of 45°, at which point it is released. At the very bottom of its swing it strikes a 500 g steel block resting on a frictionless table. To what angle does the ball rebound?

mA(v1x)A + mB(v1x)B = mA(v2x)A + mB(v2x)B

(v1x)A = 2.40 m/s(v1x)B = 0 m/s

(v2x)A = ?(v2x)B = ?

A B BA

0.48 + 0 = 0.20(v2x)A + 0.50(v2x)B

½mA(v1)A2 + ½mB(v1)B

2 = ½mA(v2)A2 + ½mB(v2)B

2

0.57 + 0 = 0.10(v2)A2 + 0.25(v2)B

2

(2)

2 A2 B

0.48 0.200.50

xx

vv

substitute (1) into (2)…

(lots of algebra)(v2)A = –1.03 m/s (or 2.40 m/s)

(1)

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A 200 g steel ball on a 1 m-long string is pulled sideways until the string is at an angle of 45°, at which point it is released. At the very bottom of its swing it strikes a 500 g steel block resting on a frictionless table. To what angle does the ball rebound?

(v2)A = –1.03 m/s(y2)A = 0 m

A

(v3)A = 0 m/s(y3)A = ?

½mA(v3)A2 + mAg(y3)A = ½mA(v2)A

2 + mAg(y2)A

0 + mAg(y3)A = ½mA(1.03) 2 + 0

(y3)A = 0.054 m

3 = 18.9°

3 = ?

0

L = 1 m

y3

y

BA = L(1 – cos3)

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PHY1012F

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ENERGYLearning outcomes:

At the end of this chapter you should be able to…Define and use the concepts of kinetic and potential energy.Use various pictorial representations to interpret problems involving the transformation of energy from one kind to another. Apply the law of conservation of mechanical energy to solve mechanical problems in conservative systems, including (in conjunction with the law of conservation of momentum) problems involving elastic collisions.