Physics 111. Lecture 15 (Walker: 7.1-2) Work & Energy March 2, Wednesday - Midterm 1

Transcription

1 Physics 111 Lecture 15 (Walker: 7.1-2) Work & Energy March 2, 2009 Wednesday - Midterm 1 Lecture 15 1/25 Work Done by a Constant Force The definition of work, when the force is parallel to the displacement: (7-1) SI work unit: newton-meter (N m) = joule, J Lecture 15 2/25

2 Typical Work Amounts Lecture 15 3/25 Work for Force at an Angle If the force is at an angle to the displacement: (7-3) Only the horizontal component of the force does any work (horizontal displacement). Lecture 15 4/25

3 Work Summary Energy is transferred from person to spring as the person stretches spring. This is work. W = F x Work = 0 W = F x= Fcosθ x x SI Unit for work: 1 joule = 1 J = 1 N m Lecture 15 5/25 Work Done by Gravity W = Fd cosθ = mgd cosθ What is total work done by gravity over total length of slide? Lecture 15 6/25

4 Negative and Positive Work The work done may be positive, zero, or negative, depending on the angle between the force and the displacement: Lecture 15 7/25 Perpendicular Force and Work Car is traveling on a curved highway. Force due to friction f s points toward center of circular path. How much work does the frictional force do on the car? Zero! General Result: A force that is everywhere perpendicular to the motion does no work. Lecture 15 8/24

5 Work on System with Many Forces W = F x + F x + F x +L total 1x 1 2x 2 3x 3 Model the system as a particle total 1x 2x 3x W = F x+ F x+ F x+ L a single x = ( F + F + F + L) x= F x 1x 2x 3x net x Lecture 15 9/24 Work Done by a Constant Force If there is more than one force acting on an object, we can find the work done by each force, and also the work done by the net force: (7-5) Lecture 15 10/24

6 Example: Pulling a Suitcase Rope inclined upward at 45 o pulls suitcase through airport. Tension on the rope is 20 N. How much work does the tension do if suitcase is pulled 100 m? W = T( x)cosθ o W = (20 N)(100 m)cos 45 = 1410 J Note that the same work could have been done by a tension of just 14.1 N by pulling in the horizontal direction. Lecture 15 11/25 Gravitational Work In lifting an object of weight mg by a height h, the person doing the lifting does an amount of work W = mgh. If the object is subsequently allowed to fall a distance h, gravity does work W = mgh on the object. Lecture 15 12/24

7 Example: Loading Ship 3,000 kg truck is to be loaded onto a ship by a crane that exerts an upward force of 31 kn on truck. This force is applied over a distance of 2.0 m. (a) Find work done on truck by crane. (b) Find work done on truck by gravity. (c) Find net work done on the truck. Wapp = Fapp y y = (31 kn)(2.0 m) = 62 kj W mg y 2 g = y = (3000 kg)( 9.81 m/s )(2.0 m) = 58.9 kj Wnet = Wapp + Wg = (62.0 kj) + ( 58.9 kj) = 3.1 kj Lecture 15 13/25 Positive & Negative Gravity Work When positive work done on an object, its speed increases; when negative work is done, its speed decreases. Lecture 15 14/25

8 Kinetic Energy & The Work-Energy Theorem After algebraic manipulations of the equations of motion, we find: Therefore, we define the kinetic energy K: (7-6) Lecture 15 15/25 Kinetic Energy & The Work-Energy Theorem Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy. (7-7) Lecture 15 16/25

9 Question 1 Car 1 has twice the mass of Car 2, but they both have the same kinetic energy. If the speed of Car 1 is v, approximately what is the speed of Car 2? a) 0.50 v b) v c) v d) v e) 2.00 v Lecture 15 17/25 Problem Solving Strategy - Work 1. Draw particle (as dot) at its initial and final positions. Label initial and final positions. 2. Put one or more coordinate axes on the drawing. 3. Draw arrows for initial and final velocities, and label them. 4. On the initial-position drawing, place a labeled vector for each force acting on object. 5. Calculate the total work done on the particle by the forces and equate this total to the change in the particle s kinetic energy. Check: Pay attention to negative signs. Values for work done can be positive or negative, depending on the direction of the displacement relative to the direction of the force. Kinetic energy values are always positive. Lecture 15 18/25

10 Example: A Dogsled Race You pull a sled (mass 80 kg) with force of 180 N at 40 above the horizontal. Sled moves x = 5.0 m, starting from rest. Neglect friction. (a) Find the work you do. (b) Find final speed of sled. Lecture 15 19/25 W = W = F x= F θ x total you x cos = (180 N)(cos 40 )(5.0 m) = 689 J W = mv mv = mv v v total 2 f 2 i 2 f 2W = m 2 total f 2W 2(689 J) m (80 kg) total f = = = 4.15 m/s Lecture 15 20/25

11 Example: Work & KE-Rocket Launch 150,000 kg rocket launched straight up. Rocket engine generates thrust of 4.0 x 10 6 N. Rocket s speed at height 500 m? (Ignore air resistance and mass loss due to burned fuel.) Lecture 15 21/25 W F y = 6 9 thrust thrust ( ) = ( N)(500 m) = J W = w y = mg y = = 4 2 grav ( ) ( ) ( kg)(9.80 m/s )(500 m) = x 10 9 J K = mv 0 = W 2 thrust + Wgrav = J 2 K v = = 1.30 m/s m Lecture 15 22/25

12 Example: Pushing a Puck 0.5 kg ice hockey puck slides across frictionless ice with an initial speed of 2.0 m/s. Compressed air gun exerts a continuous force of 1.0 N on the puck to slow it down as it moves 0.50 m. Air gun is aimed at the front edge of puck, with the compressed air flow 30 o below the horizontal. Puck s final speed? Lecture 15 23/25 o W = F( r)cos θ = (1.0 N)(0.5 m)cos150 = J K = mv mv = W W 2 2( J) v1 = v0 + = (2.0 m/s) + = 1.51 m/s m (0.5 kg) Lecture 15 24/25

13 End of Lecture 15 Before Friday, read Walker Homework Assignment #7a should be submitted using WebAssign by 11:00 PM on Friday, March 6. Wednesday -- Midterm 1 Lecture 15 25/25