PHYSICS 220 Lecture 9 Work and Kinetic Energy

PHYSICS 220 Lecture 9 Work and Kinetic Energy Lecture 9 Purdue University, Physics 220 1

For this example: vf 2 = v 02 + 2 a x = 2 F x/m mv 2/2 = F x = W (work) [N]*[m] = J (joule) Lecture 9 Purdue University, Physics 220 2

Work and Energy • Work: Transfer of Energy by Force W = |F| | r| cos W depends on the direction of the force relative to the displacement Lecture 9 Purdue University, Physics 220 3

Work by Constant Force • Only component of force parallel to direction of motion does work! F Ø W = F r cos F WF > 0: 0< < 90 : cos( ) > 0 r F F WF = 0: =90 : cos( ) =0 r WF < 0: 90< < 270 : cos( ) < 0 r r F Lecture 9 WF > 0: 0< < 90 : cos( ) > 0 Purdue University, Physics 220 4

Question You are towing a car up a hill with constant velocity. The work done on the car by the normal force is: A) positive B) negative C) zero N V T Normal force is perpendicular to direction of displacement, so work is zero. Lecture 9 Purdue University, Physics 220 W 5

Question You are towing a car up a hill with constant velocity. The work done on the car by the gravitational force is: A) positive B) negative C) zero N V T Gravity is pushing against the direction of motion so it is negative. Lecture 9 Purdue University, Physics 220 W 6

Question You are towing a car up a hill with constant velocity. The work done on the car by the tension force is: A) positive B) negative C) zero N V T The force of tension is in the same direction as the motion of the car, making the work positive. Lecture 9 Purdue University, Physics 220 W 7

i. Clicker You toss a ball in the air. The work done by gravity as the ball goes up is: A) Positive Lecture 9 B) Negative C) Zero Purdue University, Physics 220 8

Work by Constant Force • Example: You pull a 30 N chest 5 meters across the floor at a constant speed by applying a force of 50 N at an angle of 30 degrees. How much work is done by the 50 N force? W = T x cos = (50 N) (5 m) cos (30) = 217 Joules N T f mg 50 N 30 Lecture 9 Purdue University, Physics 220 9

Where did the Energy go? • Example: You pull a 30 N chest 5 meters across the floor at a constant speed, by applying a force of 50 N at an angle of 30 degrees. • How much work did gravity do? W = mg r cos = 30 5 cos(90) = 0 • How much work did friction do? r 90 mg x-direction: SF = ma T cos(30) – f = 0 f = T cos(30) W = f r cos(180) = 50 cos(30) 5 cos(180) = -217 Joules Lecture 9 Purdue University, Physics 220 N T f mg f r 180 10

Work by Variable Force Lecture 9 Purdue University, Physics 220 11

Work by Variable Force Spring: F spring= -k x Work is the area under the F vs x plot W by spring force = -1/2 k x 2 Lecture 10 Purdue University, Physics 220 12

Kinetic Energy: Motion • Apply constant force along x-direction to a point particle m W = Fx x = m ax x = ½ m (vf 2 – v 02) • Work changes ½ m v 2 • Define Kinetic Energy (energy of motion) KE = ½ m v 2 W = KE Lecture 9 Work-Kinetic Energy Theorem Purdue University, Physics 220 13

Falling Ball Example • Ball falls a distance of 5 meters. What is its final speed? Only force/work done be gravity Wg = m ½ (vf 2 – vi 2) Fg h = ½m vf 2 mgh = ½m v f 2 mg Vf = sqrt( 2 g h ) = 10 m/s Lecture 9 Purdue University, Physics 220 14

Example: Block with Friction • A block is sliding on a surface with an initial speed of 5 m/s. If the coefficient of kinetic friction between the block and table is 0. 4, how far does the block travel before stopping? N y-direction: F=ma N-mg = 0 N = mg Work WN = 0 Wmg = 0 Wf = f x cos(180) = -mmg x 5 m/s Lecture 9 f y x mg W = KE -mmg x = ½ m (vf 2 – v 02) -mg x = ½ (0 – v 02) mg x = ½ v 02 / mg = 3. 1 meters Purdue University, Physics 220 15

Work: Energy Transfer due to Force • Force to lift trunk at constant speed – Case a Ta – mg = 0 Ta = mg – Case b 2 Tb - mg =0 or Tb = ½ mg • But in case b, trunk only moves ½ distance you pull rope • Work is same in both! Ta Lecture 9 mg Tb Tb W = mgh mg Purdue University, Physics 220 16

i. Clicker A box is pulled up a rough (m > 0) incline by a ropepulley-weight arrangement as shown below. How many forces are doing (non-zero) work on the box? A) 0 Lecture 9 B) 1 C) 2 Purdue University, Physics 220 D) 3 E) 4 17

Solution Draw FBD of box: T l Consider direction of motion of the box l Any force not perpendicular to the motion will do work: v N f N does no work (perp. to v) T does positive work f does negative work 3 forces do work mg mg does negative work Lecture 9 Purdue University, Physics 220 18