8.4 Changes in Mechanical Energy for Nonconservative Forces

Mechanical Energy & Nonconservative Forces If several forces act (conservative & nonconservative): The total work done is: Wnet = WC + WNC WC = work done by conservative forces WNC = work done by non-conservative forces The work energy principle still holds: Wnet = K For conservative forces (definition of U):WC = -U  K = -U + WNC  WNC = K + U Work done by Nonconservative Forces = Total change in K + Total change in U

Mechanical Energy & Nonconservative Forces, 2 The work done against friction is greater along the red path than along the blue path Because the work done depends on the path, friction is a nonconservative force

Mechanical Energy & Nonconservative Forces, 3 If the work done against the kinetic friction depends on the path taken. As the moves moves from A to B. W ≡ – ƒkd (8.13) d is greater for the curve path so W is greater. The work done by the nonconservative force ƒk is: WNC ≡ –ƒkd Since: WNC = K + U 

Mechanical Energy & Nonconservative Forces, final In general, if friction is acting in a system: DEmech = DK + DU = –ƒkd (8.14) DU is the change in all forms of potential energy If friction is zero, this equation becomes the same as Conservation of Mechanical Energy

Problem Solving Strategies – Nonconservative Forces Define the isolated system and the initial and final configuration of the system Identify the configuration for zero potential energy These are the same as for Conservation of Energy The difference between the final and initial energies is the change in mechanical energy due to friction

Example 8. 6 Motion on a Curve Track No-conservative Forces (Example 8 Example 8.6 Motion on a Curve Track No-conservative Forces (Example 8.7 Text Book) A child of mass m starts sliding from rest. Frictionless! Find speed v at the bottom. DEmech = DK + DU DEmech =(Kf – Ki) + (Uf – Ui) = 0 (½mvf2 – 0) + (0 – mgh)= 0 ½mvf2 – mgh = 0  Same result as the child is falling vertically trough a distance h!

Example 8.6 Motion on a Curve Track, final If a kinetic friction acts on the child, find DEmech Assuming: m = 20.0 kg and vf = 3.00m/s DEmech = (Kf – Ki) + (Uf – Ui) ≠ 0 DEmech = ½mvf2 – mgh DEmech = ½(20)(3)2 – 20(9.8)(2) = –302J If we want to find k: DEmech = –302J DEmech = –ƒkd = –knd = –302J  k= 302/nd

Example 8. 7 Spring-Mass Collision No-conservative Forces (Example 8 Example 8.7 Spring-Mass Collision No-conservative Forces (Example 8.9 Text Book) Frictionless! K +Us = Emech remains constant Assuming: m= 0.80kg vA = 1.2m/s k = 50N/m Find maximum compression of the spring after collision (xmax) EC = EA  KC + UsC = KA + UsA ½mvC2 + ½kxmax2 = ½mvA2 + ½kxA2 0 + ½kxmax2 = ½mvA2 + 0 

Example 8.7 Spring-Mass Collision, 2 If friction is present, the energy decreases by: DEmech = –ƒkd Assuming: k= 0.50 m= 0.80kg vA = 1.2m/s k = 50N/m Find maximum compression of the spring after collision xC DEmech = –ƒk xC = –knxC DEmech = –kmgxC  DEmech = –3.92xC (1)

Example 8.7 Spring-Mass Collision, final Using: DEmech = Ef – Ei DEmech = (Kf – Uf) + (Ki – Ui) DEmech = 0 – ½kxC2 + ½mvA2 + 0 DEmech = – 25xC2 + 0.576 (2) Taking: (1) = (2) – 25xC2 + 0.576 = –3.92xC Solving the quadratic equation for xC : xC = 0.092m < 0.15m (frictionless) Expected! Since friction retards the motion of the system xC = – 0.25m Does not apply since the mass must be to the right of the origin.

Example 8.8 Connected Blocks in Motion N-nconservative Forces (Example 8.10 Text Book) The system consists of the two blocks, the spring, and Earth. Gravitational and potential energies are involved System is released from rest when spring is unstretched. Mass m2 falls a distance h before coming to rest. Find k

Example 8.8 Connected Blocks in Motion, 2 The kinetic energy is zero if our initial and final configurations are at rest Block 2 undergoes a change in gravitational potential energy The spring undergoes a change in elastic potential energy Emech = K + Ug + US Emech = Ug + US Emech = Ugf – Ugf + Usf – Usi Emech = 0 – m2gh + ½kh2 – 0 Emech = – m2gh + ½kh2 (1)

Example 8.8 Connected Blocks in Motion, final If friction is present, the energy decreases by: DEmech = –ƒkh = – km1gh (2) Taking (1) = (2) – m2gh + ½kh2 = – km1gh Solving for k km1gh = m2gh – ½kh2  k = (m2gh)/ (m1gh) – (½kh2)/(m1gh)  k = m2/m1 – (½kh)/m1g This is another way to measure k !!!

We will not cover Section: 8.5 Please Read it!!! Here you have a taste 8.5 Conservative Forces and Potential Energy NOTE: We will not cover Section: 8.5 Please Read it!!! Here you have a taste

Conservative Forces and Potential Energy Define a potential energy function, U, such that the work done by a conservative force equals the decrease in the potential energy of the system The work done by such a force, F, is (8.15) DU is negative when F and x are in the same direction

Conservative Forces and Potential Energy The conservative force is related to the potential energy function through (8.15) The x component of a conservative force acting on an object within a system equals the negative of the potential energy of the system with respect to x

Conservative Forces and Potential Energy – Check Look at the case of a deformed spring This is Hooke’s Law

8.6 Energy Diagrams and Equilibrium Motion in a system can be observed in terms of a graph of its position and energy In a spring-mass system example, the block oscillates between the turning points, x = ±xmax The block will always accelerate back toward x = 0

Energy Diagrams and Stable Equilibrium The x = 0 position is one of stable equilibrium Configurations of stable equilibrium correspond to those for which U(x) is a minimum x =xmax and x = –xmax are called the turning points

Energy Diagrams and Unstable Equilibrium Fx = 0 at x = 0, so the particle is in equilibrium For any other value of x, the particle moves away from the equilibrium position This is an example of unstable equilibrium Configurations of unstable equilibrium correspond to those for which U(x) is a maximum

Neutral Equilibrium Neutral equilibrium occurs in a configuration when over some region U is constant A small displacement from a position in this region will produce either restoring or disrupting forces

Potential Energy in Molecules There is potential energy associated with the force between two neutral atoms in a molecule which can be modeled by the Lennard-Jones function

Potential Energy Curve of a Molecule Find the minimum of the function (take the derivative and set it equal to 0) to find the separation for stable equilibrium The graph of the Lennard-Jones function shows the most likely separation between the atoms in the molecule (at minimum energy)

Force Acting in a Molecule The force is repulsive (positive) at small separations The force is zero at the point of stable equilibrium The force is attractive (negative) when the separation increases At great distances, the force approaches zero

Material for the Final Examples to Read!!! Example 8.6 (page 230) Homework to be solved in Class!!! Questions: 13, 21 Problems: 36, 45, 48