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Extended - Theorem

 We denote the total by

E  KE  PE  PEs

 Since (KE  PE  PEs ) f  (KE  PE  PEs )i

 The total mechanical energy is conserved and remains the same at all 1 1 1 1 mv 2  mgy  kx2  mv 2  mgy  kx2 2 i i 2 i 2 f f 2 f A block projected up a incline

 A 0.5-kg block rests on a horizontal, frictionless surface. The block is pressed back against a spring having a constant of k = 625 N/m, compressing the spring by 10.0 cm to point A. Then the block is released.  (a) Find the maximum d the block travels up the frictionless incline if θ = 30°.  (b) How fast is the block going when halfway to its maximum height? A block projected up a incline  Point A (initial state): v  0, y  0, x  10cm  0.1m  Point B (final state): i i i v f  0, y f  h  d sin , x f  0 1 1 1 1 mv 2  mgy  kx2  mv 2  mgy  kx2 2 i i 2 i 2 f f 2 f 1 kx2  mgy  mgd sin  2 i f

1 kx2 0.5(625N / m )( 0.1 m )2 d 2 i mg sin (0.5kg )(9.8 m / s2 )sin30 1.28m A block projected up a incline

 Point A (initial state): vi  0, yi  0, xi  10cm  0.1m

 Point B (final state): v f  ?, y f  h / 2  d sin  / 2, x f  0 1 1 1 1 mv 2  mgy  kx2  mv 2  mgy  kx2 2 i i 2 i 2 f f 2 f

1 2 1 2 h k 2 2 kxi  mv f  mg ( ) x  v  gh 2 2 2 m i f h  d sin   (1.28m)sin 30   0.64m k v  x2  gh f m i  ......  2.5m / s Types of  Conservative forces  Work and energy associated with the can be recovered  Examples: , Spring Force, EM forces  Nonconservative forces  is often dispersed and work done against it cannot easily be recovered  Examples: Kinetic , air forces, normal forces, tension forces, applied forces … Conservative Forces

 A force is the work of an object moving between two points, if the object is independent of the path between the points

 The work depends only on the starting and ending positions of the object

 Any can have a function associated with it

 Work done by gravity Wg  PEi  PE f  mgyi  mgy f  Work done by spring force 1 1 W  PE  PE  kx2  kx2 s si sf 2 i 2 f Nonconservative Forces

 If the work you do on an object depends on the path between the final and the starting points of the object, then a power is not conserved.  Work depends on the path of  For a non-conservative force, the potential energy can not be identifiedWork done by a nonconservative force   Wnc  F d   fk d  Wotherforces Figure 8.10  It is generally dissipative. The dispersal Physics for Scientists and of energy takes the form of or soundEngineers 6th Edition, Thomson Brooks/Cole © 2004; Chapter 8

Physics for Scientists and Engineers 6th Edition, Thomson Brooks/Cole © 2004; Extended Work-Energy Theorem

 The work-energy theorem can be written as:

WKEKEKEnet f  i  

Wnet Wnc Wc

 Wnc represents the work done by nonconservative forces  Wc represents the work done by conservative forces  Any work done by conservative forces can be accounted for by changes in potential energy Wc  PEi  PE f

 Gravity work Wg  PEi  PE f  mgyi  mgy f  Spring force work 1 1 W  PE  PE  kx2  kx2 s i f 2 i 2 f Extended Work-Energy Theorem

 Any work done by conservative forces can be accounted for by changes in potential energy

Wc  PEi  PE f  (PE f  PEi )  PE

Wnc  KE  PE  (KE f  KEi )  (PE f  PEi )

Wnc  (KE f  PE f )  (KEi  PEi )

 Mechanical energy include kinetic and potential energy 1 1 E  KE  PE  KE  PE  PE  mv2  mgy  kx2 g s 2 2

Wnc  E f  Ei Conservation of Mechanical Energy  A block of m = 0.80 kg slides across a horizontal frictionless counter with a of v = 0.50 m/s. It runs into and compresses a spring of spring constant k = 1500 N/m. When the block is momentarily stopped by the spring, by what distance d is the spring compressed?

Wnc  (KE f  PE f )  (KEi  PEi )

1 1 1 1 1 2 1 2 mv 2  mgy  kx2  mv 2  mgy  kx2 0 0 kd  mv  0  0 2 f f 2 f 2 i i 2 i 2 2

1 1 0 0 kd2  mv 2  0  0 2 2 m d  v2 1.15cm k

Physics for Scientists and Engineers 6th Edition, Thomson Brooks/Cole © 2004; Changes in Mechanical Energy for conservative forces

 A 3-kg crate slides down a ramp. The ramp is 1 m in length and inclined at an of 30° as shown. The crate stats from rest at the top. The surface friction can be negligible. Use energy methods to determine the speed of the crate at the bottom of the ramp. 1 1 1 1  fd  W  ( mv 2  mgy  kx2 ) ( mv 2  mgy  kx2 )  otherforces 2 f f 2 f 2 i i 2 i 1 1 1 1 ( mv 2  mgy  kx2 )  ( mv 2  mgy  kx2 ) 2 f f 2 f 2 i i 2 i

 d 1m, yi  d sin 30  0.5m,vi  0 y  0,v  ? Figure 8.11 f f Physics for Scientists and Engineers 6th Edition, 1 2 ( mv f  0  0)  (0  mgyi  0) Thomson Brooks/Cole © 2 2004; Chapter 8

v f  2gyi  3.1m/ s

Physics for Scientists and Engineers 6th Edition, Thomson Brooks/Cole © 2004; Changes in Mechanical Energy for Non-conservative forces

 A 3-kg crate slides down a ramp. The ramp is 1 m in length and inclined at an angle of 30° as shown. The crate stats from rest at the top. The surface in contact have a coefficient of kinetic friction of 0.15. Use energy methods to determine the speed of the crate at the bottom of the ramp. 1 1 1 1  fd  W  ( mv 2  mgy  kx2 ) ( mv 2  mgy  kx2 )  otherforces 2 f f 2 f 2 i i 2 i 1   Nd  0  ( mv 2  0 0) (0 mgy  0) k 2 f i  k  0.15,d 1m, yi  d sin 30  0.5m, N  ? N  mg cos  0 Figure 8.11 1   dmg cos  mv 2  mgy Physics for Scientists and k 2 f i Engineers 6th Edition, Thomson Brooks/Cole © v f  2g(yi  k d cos)  2.7m/ s 2004; Chapter 8 Changes in Mechanical Energy for Non-conservative forces

 A 3-kg crate slides down a ramp. The ramp is 1 m in length and inclined at an angle of 30° as shown. The crate stats from rest at the top. The surface in contact have a coefficient of kinetic friction of 0.15. How far does the crate slide on the horizontal floor if it continues to experience a friction force. 1 1 1 1  fd  W  ( mv 2  mgy  kx2 ) ( mv 2  mgy  kx2 )  otherforces 2 f f 2 f 2 i i 2 i 1   Nx  0  (0 0 0) ( mv 2  0 0) k 2 i   0.15,v  2.7m/ s, N  ? k i Figure 8.11 N  mg  0 Physics for Scientists and Engineers 6th Edition, 1 Thomson Brooks/Cole ©   mgx   mv 2 k 2 i 2004; Chapter 8 v2 x  i  2.5m 2k g Block-Spring Collision

 A block having a mass of 0.8 kg is given an initial vA = 1.2 m/s to the right and collides with a spring whose mass is negligible and whose force constant is k = 50 N/m as shown in figure. Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision. 1 1 1 1 mv 2  mgy  kx2  mv 2  mgy  kx2 2 f f 2 f 2 i i 2 i

1 2 1 2 Figure 8.14 mv max  0  0  mv A  0  0 Physics for Scientists and 2 2 Engineers 6th Edition, Thomson Brooks/Cole © m 0.8kg 2004; Chapter 8 x  v  (1.2m/ s)  0.15m max k A 50 N / m Block-Spring Collision

 A block having a mass of 0.8 kg is given an initial velocity vA = 1.2 m/s to the right and collides with a spring whose mass is negligible and whose force constant is k = 50 N/m as shown in figure. Suppose a constant force of kinetic friction acts between the block and the

surface, with µk = 0.5, what is the maximum compression xc in the spring. 1 1 1 1  fd  W  ( mv 2  mgy  kx2 ) ( mv 2  mgy  kx2 )  otherforces 2 f f 2 f 2 i i 2 i 1 1   Nd  0  (0 0 kx2 ) ( mv 2  0 0) k 2 c 2 A

N  mg and d  xc 1 1 kx2  mv 2   mgx 2 c 2 A k c

2 25 xc  3.9xc  0.58  0 xc  0.093m Energy Review

 Associated with movement of members of a system  Potential Energy  Determined by the configuration of the system  Gravitational and Elastic   Related to the of the system  Energy is conserved  This means that energy cannot be created nor destroyed  If the total amount of energy in a system changes, it can only be due to the fact that energy has crossed the boundary of the system by some method of energy transfer Practical Case

  E =  K +  U = 0

 The total amount of energy in the system is constant. 1 1 1 1 mv 2  mgy  kx2  mv 2  mgy  kx2 2 f f 2 f 2 i i 2 i Practical Case

  K +  U + Eint = W + Q + TMW + TMT + TET + TER

 The Work-Kinetic Energy theorem is a special case of Conservation of Energy  K +  U = W Ways to Transfer Energy Into or Out of A System

 Work – transfers by applying a force and causing a of the point of application of the force  Mechanical Waves – allow a disturbance to propagate through a medium  Heat – is driven by a temperature difference between two regions in  Matter Transfer – matter physically crosses the boundary of the system, carrying energy with it  Electrical Transmission – transfer is by  Electromagnetic – energy is transferred by electromagnetic waves Connected Blocks in Motion  Two blocks are connected by a string that passes over a frictionless pulley. The block of mass m1 lies on a horizontal surface and is connected to a spring of force constant k. The system is released from rest when the spring is unstretched. If the hanging block of mass m2 fall a distance h before coming to rest, calculate the coefficient of kinetic friction between the block of mass m1 and the surface.  fd  Wotherforces  KE  PE

1 2 PE  PE g  PEs  (0 m2 gh)  ( kx 0) 2 Figure 8.15 1 2  k Nx  0  m2 gh kx Physics for Scientists and 2 Engineers 6th Edition, N  mg and x  h Thomson Brooks/Cole © 2004; Chapter 8 1   m gh  m gh kh2 1 k 1 2 2 m2 g  kh Physics for Scientists and Engineers 6th Edition, Thomson 2 Brooks/Cole © 2004; k  m1g