Dynamics Energy and Work

Transcription

1 Dynamics Energy and Work Lana Sheridan De Anza College Oct 24, 2017

2 Last Time resistive forces: Drag Equation

3 Drag Equation, One more point What if the object is not dropped from rest? (See Ch 6, prob 33, for example) If an object starts out with a large velocity, then begins to experience an unbalanced resistive force it will have an acceleration opposite to its direction of motion. Consider what happens during a skydive.

4 Overview Energy (systems and environments) Work

5 Energy Energy is almost impossible to clearly define, yet everyone has a good intuitive notion of what it is. Energy is a property of physical systems. It tells us something about the states or configurations the system can be in. In fact, it is possible to find the dynamics of a system purely from understanding the distribution of energy in the system.

6 Energy Energy is almost impossible to clearly define, yet everyone has a good intuitive notion of what it is. Energy is a property of physical systems. It tells us something about the states or configurations the system can be in. In fact, it is possible to find the dynamics of a system purely from understanding the distribution of energy in the system. Importantly, it can neither be created or destroyed, but it can be transferred between systems, and take different forms.

7 Types of Energy motional energy (kinetic) energy as a result of object s configurations or stored energy (potential) heat, light, sound - can carry away energy from a mechanical system

8 Systems and Environments To make some predictions about physical objects, you must identify a system of interest. Some object or collection of objects. System

9 Systems and Environments To make some predictions about physical objects, you must identify a system of interest. Some object or collection of objects. System They are modeled. The model may not be exactly accurate, just good enough to make the predictions you want.

10 Systems and Environments To make some predictions about physical objects, you must identify a system of interest. Some object or collection of objects. System They are modeled. The model may not be exactly accurate, just good enough to make the predictions you want. Outside influences (eg. forces or energies) can be included in the description, but the source of these effects is not described. Everything outside the system is the system s environment.

11 does not move rse, Work we apply a gest that when, we must cone of the force. nt represents a N through the rtant. Moving ving it 2 cm if with the system. After on, insert the system. For example, the work done by the hammer on the nail Let s consider how identifies the environment the nail as can the system, effect the system by and the force from the hammer exchanging energy with it. represents the influence from the Take the systemenvironment. to be a block. An external force (from the environment) acts on it. system) undert force of magt. r orce on the itude Dr of Figure 7.2 An object undergoes This force affects a the displacement block: it can D S r under accelerate the it. u, where u is action of a constant force S F. It can also change the energy of the block. u S F S

12 N through the rtant. WorkMoving ving it 2 cm if system) undert force of magt. r orce on the itude Dr of Figure 7.2 An object undergoes For a constant applied force Work is defined as: u, where u is W = F r (7.1) Work done by a Work is the amount constant of energy force transferred to a system by an fined interaction terms with the environment. xplore Units: how Joules, to J. 1 J = 1 Nm t of application u S F S a displacement D r S under the action of a constant force F S.

13 Vectors W 5 F Dr Properties cos u5 S F? D and Operations S r (7.3) Multiplication by a vector: S r is a shorthand notation for F Dr cos u. with our discussion of work, let us investigate some properties igure 7.6 The shows two Dot vectors Product S A and S B and the angle u between inition of the dot product. In Figure 7.6, B cos u is the projecerefore, Let Equation A = 7.2 A x means i + Athat y j S A? S B is the product of the the projection of B = S B onto B x i + S A. B 1 y j, nd side of Equation 7.2, we also see that the scalar product is is, S S S S A? B 5 B? A A B = A x B x + A y B y The output of this operation is a scalar. to stating that A S? B S equals the product of the magnitude of B S and the projection of A S Equivalently, nother way of combining vectors that proves useful in physics and is not commutative. A B = AB cos θ u B cos u S B S A. = AB cos u B S S A Figure 7.6 The scalar product S S S A? B equals the magnitude of A multiplied by B cos u, which is the projection of S B onto S A.

14 Vectors W 5 F Dr Properties cos u5 S F? D and Operations S r (7.3) Multiplication by a vector: S r is a shorthand notation for F Dr cos u. with our discussion of work, let us investigate some properties igure 7.6 The shows two Dot vectors Product S A and S B and the angle u between inition of the dot product. In Figure 7.6, B cos u is the projecerefore, Let Equation A = 7.2 A x means i + Athat y j S A? S B is the product of the the projection of B = S B onto B x i + S A. B 1 y j, nd side of Equation 7.2, we also see that the scalar product is is, S S S S A? B 5 B? A A B = A x B x + A y B y The output of this operation is a scalar. to stating that A S? B S equals the product of the magnitude of B S and the projection of A S Equivalently, nother way of combining vectors that proves useful in physics and is not commutative. Properties A B = AB cos θ The dot product is commutative: A B = B A If A B, A B = AB. If A B, A B = 0. u B cos u S B S A. = AB cos u B S S A Figure 7.6 The scalar product S S S A? B equals the magnitude of A multiplied by B cos u, which is the projection of S B onto S A.

15 Vectors Properties and Operations Multiplication by a vector: The Dot Product Try it! Find A B when A is a vector of magnitude 6 N directed at 60 above the x-axis and B is a vector of magnitude 2 m pointed along the x-axis.

16 Vectors Properties and Operations Multiplication by a vector: The Dot Product Try it! Find A B when A is a vector of magnitude 6 N directed at 60 above the x-axis and B is a vector of magnitude 2 m pointed along the x-axis. (1 J = 1 Nm) A B = (6 N)(2 m) cos(60 ) = 6 J

17 Vectors Properties and Operations Multiplication by a vector: The Dot Product Try it! Find A B when A is a vector of magnitude 6 N directed at 60 above the x-axis and B is a vector of magnitude 2 m pointed along the x-axis. (1 J = 1 Nm) Now find A B when: A B = (6 N)(2 m) cos(60 ) = 6 J A = 1 i + 2 j ; B = 1 i 4 j

18 Vectors Properties and Operations Multiplication by a vector: The Dot Product Try it! Find A B when A is a vector of magnitude 6 N directed at 60 above the x-axis and B is a vector of magnitude 2 m pointed along the x-axis. (1 J = 1 Nm) Now find A B when: A B = (6 N)(2 m) cos(60 ) = 6 J A = 1 i + 2 j ; B = 1 i 4 j A B = (1)( 1) + (2)( 4) = 9

19 Work If there are several forces acting on a system, each one can have an associated work. In other words, we can ask what is the work done on the system by each force separately. W i = F i ( r)

20 Work 180 Chapter 7 Energy of a System S F is the only force that does work on the block in this situation. S n mg S u r S S F 1 Figure from Serway Figure & Jewett. 7.3 An object is dis- have done a tion, howev the chair, b does not mo the situation Also noti is zero when application Figure 7.3, the gravitat dicular to t tion of D r S. The sign done by the

21 Net Work Work done on the system by each force separately: W i = F i ( r) Net Work W net = i W i where the sum includes the work of all forces acting on the system. If the system is treated as a particle: W net = F net ( r)

22 d u Units of Work Thus, whenever we calculate work, we must be careful about its sign and assume it to be positive. Work can be positive or negative! the e- nerro gles F F F d d d 90 < < 90 = ± < < 270 W = Fd cos θ > 0 (a) (b) (c) W = Fd cos θ = 0 W = Fd cos θ < 0 positive work zero work negative work For work done on a system: Positive energy is transferred to the system. Negative energy is transferred from the system.

23 Question Quick Quiz The gravitational force exerted by the Sun on the Earth holds the Earth in an orbit around the Sun. Let us assume that the orbit is perfectly circular. The work done by this gravitational force during a short time interval in which the Earth moves through a displacement in its orbital path is (A) zero (B) positive (C) negative (D) impossible to determine 1 Serway and Jewett, page 180.

24 Question Quick Quiz The gravitational force exerted by the Sun on the Earth holds the Earth in an orbit around the Sun. Let us assume that the orbit is perfectly circular. The work done by this gravitational force during a short time interval in which the Earth moves through a displacement in its orbital path is (A) zero (B) positive (C) negative (D) impossible to determine 1 Serway and Jewett, page 180.

25 Summary Energy, systems, environments Work 2nd Collected Homework! due Friday, Oct 27. 2nd Test Friday, Nov 3. (Uncollected) Homework Serway & Jewett, Read Chapter 7. Ch 7, onward from page 204. Probs 3, 5, 7, 11, 9, 13, 15, 17