Energy present in a variety of forms. Energy can be transformed form one form to another Energy is conserved (isolated system) ENERGY

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1 ENERGY Energy present in a variety of forms Mechanical energy Chemical energy Nuclear energy Electromagnetic energy Energy can be transformed form one form to another Energy is conserved (isolated system) (riction) WORK Energy is the ability to do WORK Work done by a constant force Direction of Motion Displacement = x Work = orce x displacement W = (cos) x = angle between the force and the displacement Case 1: = W = (cos ) x = x > Case 2: = 18 W = (cos18 ) x = - x < // Case 3 (general): W = (cos ) x = // x // = component of parallel to displacement Scalar WORK Work = orce x displacement SI Units: (Newton)(meter) = Nm or Joules (J) Direction of Motion Displacement = x What is the work done by? = 9 so W = (cos9 ) x = J 1

2 Work Done by a Constant orce cos 1-1 Displacement d < < 9 W > 9 < < 18 W < Problem 8: A block of mass 2.5 kg is pushed 2.2 m along a frictionless horizontal table by a constant 16. N force directed 25 below the horizontal. N x = 2.2 m W Determine a) the work done by the applied force: W = (16. cos25 N)(2.2 m) = 31.9 Nm = 31.9 J b) the work done by the normal force: W N = (N cos9 N)(2.2 m) = J c) the work done by the force of gravity W g = (mg cos9 N)(2.2 m) = J d) The work done by the net force on the block W net = (16. cos25 N)(2.2 m) = 31.9 J or W net = W + W N + W g = = 31.9 J Problem 5: Starting from rest, a 5. kg block slides 2.5 m down a rough 3 incline. The coefficient of kinetic friction between the block and the incline is µ k =.436. Determine y a) the work done by the force of gravity The angle between the displacement and the weight is 6 N W g = (mg cos6 N)(2.5 m) = 61.3 J x f k b) the work done by the force of friction W x = Wsin3 W y = Wcos3 6 3 f k = µ k N and N = mgcos3, = 3 so f k = µ k ( mgcos3 )= 18.5 N The angle between the displacement and the friction is 18 W friction = [(18.5 N)cos18 ](2.5 m) = J x W c) the work done by the normal force The normal force is perpendicular to the displacement, so no work is done by the normal force on the block W N = J 2

3 Problem 7: k A 5. kg block is pushed 3. m up a vertical wall with constant speed by a constant force of magnitude applied at an angle of 3 with the horizontal. The coefficient of kinetic friction between the block and the wall is µ k =.3. Determine a) the work done by the applied force The applied force makes an angle of 6 with the displacement up the wall, But we need to find the magnitude of force first. y Apply Newton s laws N Σ x = N = cos3 x Σ y = sin3 = W + f k = mg + µ k (cos3 ) 3 f (sin3 - µ k cos3 )= mg = mg /(sin3 -.3cos3 ) = 2 N sin3 6 W 3 W = [(2 N)cos6 ](3. m) = 31 J b) the work done by the force of gravity N cos3 f k x Gravity makes an angle of 18 with the displacement W W g = [(49 N)cos18 ](3. m) = - 15 J c) the work done by the normal force The normal force is perpendicular to the displacement, so no work is done by the normal force on the block W N = J x Mechanical Energy Kinetic Energy (KE): Energy of Motion KE = ½ m v 2 Scalar SI Units: kgm 2 /s 2 = Joule (J) Same Units as Work! Mechanical Energy Work Energy Theorem When work is done by a net force on an object and the only change in the object is its speed, the work done is equal to the change in the object s kinetic energy W net = KE final KE initial = KE W net : work done on the system W net >, system gains energy W net <, system loses energy To find W net : ind net force and calculate its work (harder) Or calculate work done by each force and add up works (easier) 3

4 Problem 16: A.6 kg particle has a speed of 2. m/s at point A and a kinetic energy of 7.5 J at point B. What is a) Its kinetic energy at point A? KE A = ½ mv A2 = ½(.6 kg)(2. m/s) 2 = 1.2 J b) Its speed at point B? KE B = ½ mv 2 B v B2 = 2 KE B /m v B = 2(7.5 J)/(.6 kg) = 5. m/s b) the total work done on the particle as it moves from A to B? Wnet = KE = KE B -KE A Wnet =( ) J = 6.3 J Mechanical Energy Potential Energy (PE): Energy of Position Energy associated with the position of an object within some system Property of a system not just object Gravitational Potential Energy System considered: Earth + object near surface Energy associated with relative position of an object near Earth s surface Gravitational Potential Energy: PE = mgy (weight x height) SI Units: kgm 2 /s 2 = Joules (J) Same as Work! y i = h, PE i = mgh PE = PE f PE i = - mgh y f =, PE f = Ball dropped from initial height h Ball loses potential energy 4

5 Ball thrown upward y f = h, PE f = mgh PE = PE f PE i = +mgh y i =, PE i = Ball gains potential energy References Levels for Gravitational Potential Energy Zero reference is arbitrary Change in PE does not depend on the reference level 3. m B m = 1.5 kg moves from A to B (1) PE i = mgy i = (1.5 kg)(9.8 m/s2)(2. m) = 29 J 1. m B PE f = mgy f = (1.5 kg)(9.8 m/s2)(3. m) = 44 J 2. m A PE = PE f PE i = 44 J 29 J = 15 J A (2) PE i = mgy i = (1.5 kg)(9.8 m/s2)( m) = J PE f = mgy f = (1.5 kg)(9.8 m/s2)(1. m) = 15 J PE = PE f PE i = 15 J J = 15 J (1) (2) Work done by the gravitational force y W =mg y y i As the object falls, the only force that does work on it is the gravitational force W = mg W gravity = (mgcos )(y i y f ) W gravity = mgy i mgy f W gravity = PE i -PE f W =mg y f (1) 5

6 Problem 21: The arm, including the hand has a mass of 7. kg. Treating the arm as if it were a Single point mass m attached to a rigid massless rod at point A, determine the work done by the deltoid muscle to raise the arm from position 1 to position 2. L = The muscle must do work against gravity to raise the arm: W muscle = - W gravity = - (mgy i mgy f ) = mgh Where h = L(1-cos3 ) W muscle = (7. kg)(9.8 m/s 2 )(.21m)(1-cos3 ) = 1.9 J 6

PHYSICS 231 INTRODUCTORY PHYSICS I

PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 6 Last Lecture: Gravity Normal forces Strings, ropes and Pulleys Today: Friction Work and Kinetic Energy Potential Energy Conservation of Energy Frictional Forces