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PHY1012F ENERGY. Gregor Leigh gregor.leigh@uct.ac.za. ENERGY. 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.

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Phy1012f energy

PHY1012FENERGY

Gregor Leighgregor.leigh@uct.ac.za


Energy

ENERGY

ENERGY

  • 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.

Learning outcomes:At the end of this chapter you should be able to…


Energy1

ENERGY

ENERGY

  • Energy…

  • 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.)


Kinetic energy and gravitational potential energy

ENERGY

KINETIC ENERGY and GRAVITATIONAL POTENTIAL ENERGY

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

  • mgsin

  • mgcos

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

  • y

s

Newton II:

  • yf

dy

  • yi

(chain rule)

ds

dy = sinds


Kinetic energy and gravitational potential energy1

ENERGY

KINETIC ENERGY and GRAVITATIONAL POTENTIAL ENERGY

… mvs dvs = –mgdy

  • y

s

  • yf

  • yi

½mvf2–½mvi2

= –(mgyf – mgyi)

½mvf2+mgyf= ½mvi2+ mgyi

i.e. Kf + Ugf = Ki + Ugi

½mv2 = K

Kinetic energy– energy due to a body’s motion.

mgy = Ug

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


Kinetic energy and gravitational potential energy2

ENERGY

KINETIC ENERGY and GRAVITATIONAL POTENTIAL ENERGY

  • Units: [kgm2/s2= joule, J]

  • Kinetic energy can never be negative.

K = ½mv2

Ug = mgy

  • Units: [kgm/s2m = kgm2/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. (Ucan be negative.) Irrespective of the chosen origin, however, Ug will always be computed as the same value.


Mechanical energy

ENERGY

MECHANICAL ENERGY

  • The 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


Conservation of mechanical energy

ENERGY

CONSERVATION OF MECHANICAL ENERGY

  • Although 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


Hooke s law

ENERGY

HOOKE’S LAW

  • The spring constant, k, is a measure of the “stiffness” of the spring.

  • The negative sign indicates a restoringforce.

  • s = L –L0is the displacement from equilibrium.

  • (Fsp)sis 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.

s > 0

(unstretched)

s < 0

  • s

0

se


Phy1012f energy

  • x

ENERGY

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

(Fnet)y = n– w = 0

(Fnet)x = Fsp– fs = 0

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

 kx– s mg = 0

 t= 4.7 s


Elastic potential energy

0

se

ENERGY

ELASTIC POTENTIAL ENERGY

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

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

si

  • k

  • vi

  • m

Newton II:

sf

(chain rule)

  • vf


Elastic potential energy1

ENERGY

ELASTIC POTENTIAL ENERGY

  • … mvs dvs = –ksds

= –½k(sf)2+ ½k(si)2

½mvf2–½mvi2

½mvf2+ ½k(sf)2 = ½mvi2+ ½k(si)2

½k(s)2 = Usp

Elastic potential energy– energy due to a body’s position.

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


Elastic potential energy2

ENERGY

ELASTIC POTENTIAL ENERGY

  • Units: [(N/m)(m2) = kgm2/s2= joule, J]

  • Since sis squared, elastic potential energy is positive irrespective of whether the spring is extended or compressed.

½k(s)2 = Usp

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


Vertical springs

ye

0

ENERGY

VERTICAL SPRINGS

  • 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…

K Ug Usp


Elastic collisions

ENERGY

ELASTIC COLLISIONS

  • An idealised collision in which mechanical energy is conserved is called an perfectly elastic collision.

Problem-solving strategy for elastic collisions:

  • Sketch the situation.

  • 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).

  • Establish a coordinate system to match the motion.

  • Define symbols for initial and final masses and velocities, and list known information, identify desired unknowns.

  • Apply the laws of conservation of momentum and energy.


Phy1012f energy

0

A

A

A

B

B

B

B

ENERGY

  • 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°

L=1m

3=?

mA=200g

A

mB=500g

(y0)A = ?(v0)A=0m/s

(y1)A=0m(v1x)A=?

(v1x)B=0m/s

(v2x)A=?

(v2x)B=?

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

Part 1: Cons. of E

Part 3: Cons. of E

Part 2: Cons. of P and E


Phy1012f energy

ENERGY

  • 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?

= L(1 –cos0)

= 0.293 m

  • y

(y0)A = L– Lcos0

L=1m

0=45°

A

½mA(v1)A2 + mAg(y1)A = ½mA(v0)A2 + mAg(y0)A

  • y0

 ½mA(v1)A2 + 0 = 0 + mAg(y0)A

0

B

(y1)A=0m(v1)A=?

(y0)A = ?(v0)A=0m/s

= 2.40 m/s


Phy1012f energy

ENERGY

  • 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

 0.48 + 0 = 0.20(v2x)A + 0.50(v2x)B

B

B

A

A

(1)

(v1x)A=2.40 m/s

(v1x)B=0m/s

(v2x)A=?

(v2x)B=?

½mA(v1)A2 + ½mB(v1)B2 = ½mA(v2)A2 + ½mB(v2)B2

 0.57 + 0 = 0.10(v2)A2 + 0.25(v2)B2 (2)

substitute (1) into(2)…

(lots of algebra)

(v2)A = –1.03 m/s (or 2.40 m/s)


Phy1012f energy

ENERGY

  • 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?

  • y

½mA(v3)A2 + mAg(y3)A = ½mA(v2)A2 + mAg(y2)A

L=1m

3=?

 0 + mAg(y3)A = ½mA(1.03)2 + 0

  • y3

A

 (y3)A = 0.054 m

= L(1 –cos3)

0

B

A

(v2)A=–1.03 m/s

(y2)A=0m

(v3)A=0m/s

(y3)A= ?

 3 = 18.9°


Energy2

ENERGY

ENERGY

  • 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.

Learning outcomes:At the end of this chapter you should be able to…