Potential Energy and Conservation of Mechanical Energy Conservative - PDF document

Potential Energy and Conservation of Mechanical Energy • Conservative and Nonconservative Forces • Potential Energy • Conservation of Mechanical Energy • Homework 1

First Definition of Conservative and Nonconservative Forces • A force is conservative if the kinetic energy of a parti- cle on which it acts returns to its initial value after any round trip. A force is nonconservative if the kinetic energy changes. • An example of a conservative force is the force ex- erted by a spring • An example of a nonconservative force is friction 2

Second Definition of Conservative and Nonconservative Forces • A force is conservative if the work done by the force on a particle that moves through any round trip is zero. A force is nonconservative if the work done by the force on a particle that moves through any round trip is not zero. • First and second definitions are equivalent from the Work-Energy Theorem ( W = ∆ K ) 3

Third Definition of Conservative and Nonconservative Forces • A force is conservative if the work done by it on a particle that moves between two points depends only on these points and not on the path followed. A force is nonconservative if the work done by it on a particle that moves between two points depends on the path taken between these two points. • The work done against friction depends on the path taken since it is a nonconservative force 4

Potential Energy • When conservative forces are acting (e.g. spring force, gravity) ∆ K + ∆ U = 0 where U = potential energy (has meaning only for conservative forces) • This means that the sum of the kinetic and potential energies is a constant U + K = a constant • From the Work-Energy Theorem W = ∆ K = − ∆ U • For one dimensional motion � x f ∆ U = − W = − x i F ( x ) dx 5

Conservation of Mechanical Energy • Since the sum of the kinetic and potential energies is a constant when only conservative forces are acting, we can write K i + U i = K f + U f • We can also calculate the force from the potential en- ergy function F ( x ) = − dU ( x ) dx • Check  − dU ( x )   � x f � x f ∆ U = − x i F ( x ) dx = −  dx     x i dx � x f ∆ U = x i dU ( x ) dx = U ( x f ) − U ( x i ) = U f − U i 6

Conservation of Mechanical Energy (Cont’d) Consider a particle moving from A to B along the x-axis with a single conservative force acting on it ∆ U = U B − U A � x B U B = ∆ U + U A = − x A F ( x ) dx + U A We cannot assign a value to U B until we assign one to U A . Usually we choose U A = 0 . 7

Gravitational Potential Energy F g = − mg � y � y U g ( y ) = − 0 F g dy + U g (0) = − 0 ( − mg ) dy + U g (0) U g ( y ) = mgy + U g (0) Let U g (0) = 0 U g ( y ) = mgy ⇒ F g = − dU g ( y ) = − d ( mgy ) Note : = − mg dy dy 8

Potential Energy of a Spring F s ( x ) = − kx � x � x U s ( x ) = − 0 F s ( x ) dx + U s (0) = − 0 ( − kx ) dx + U s (0) U s ( x ) = 1 2 kx 2 + U s (0) U s ( x ) = 1 2 kx 2 Let U s (0) = 0 ⇒ F s ( x ) = − dU s ( x ) = − d  1   2 kx 2  = − kx Note ;   dx dx 9

Example 1 What is the change in gravitational potential energy when a 720-kg elevator moves from street level to the top of the Empire State building, 380 m above street level? 10

Example 1 Solution What is the change in gravitational potential energy when a 720-kg elevator moves from street level to the top of the Empire State building, 380 m above street level? ∆ U = U f − U i = mgy f − mgy i = mg ( y f − y i ) � 9 . 8 m/s 2 � ∆ U = (720 kg ) (380 m ) = 2 . 7 MJ 11

Example 2 The spring in a spring gun has a force constant k = 700 N/m. It is compressed 3.0 cm from its natural length, and a 0.012-kg ball is put in the barrel against it. Assuming no friction and a horizontal gun barrel, with what speed will the ball leave the gun when released? 12

Example 2 Solution The spring in a spring gun has a force constant k = 700 N/m. It is compressed 3.0 cm from its natural length, and a 0.012-kg ball is put in the barrel against it. Assuming no friction and a horizontal gun barrel, with what speed will the ball leave the gun when released? K i + U i = K f + U f 0 + 1 2 kx 2 = 1 2 mv 2 + 0 � � � k � 700 N/m � � � � � v = � mx = 0 . 012 kg (0 . 03 m ) = 7 . 25 m/s � � � 13

Homework 12 - Due Fri. Oct. 8 • Read Sections 7.1-7.2 • Answer Questions 7.1 & 7.3 • Do Problems 7.2, 7.5, & 7.9 14