11/13/15 Oregon State University PH 211, Class #211 We transfer momentum to an object by exerting a net force on it during a time interval. That is, we exert an impulse on the object: J B = ∫ F net.B (t) dt =  P B We transfer energy to an object by exerting a net force on it while it undergoes a spatial displacement in the direction of that force. That is, we do work on the object: W B = ∫ F s.net.B (s) ds =  E B

11/13/15 Oregon State University PH 211, Class #212 Also, we know what the result of an impulse looks like: J B = ∫ F net.B (t) dt =  P B =  [m B v B ] So, what does the result of work look like? W B = ∫ F s.net.B (s) ds =  [??]

11/13/15 Oregon State University PH 211, Class #213 Notice: F net.B = dP B /dt = m B (dv B /dt) = m B (dv B /ds)(ds/dt) = m B (dv B /ds)(v B ) So: F net.B ds = m B v B dv B ∫ F net.B ds = m B ∫ v B dv B ∫ F net.B ds = (1/2)m B (v B.f 2 – v B.i 2 )

11/13/15 Oregon State University PH 211, Class #214 So now we know (as before) what the result of an impulse looks like: J B = ∫ F net.B (t) dt =  P B =  [m B v B ] And we know what the result of work looks like? W B = ∫ F s.net.B (s) ds =  [(1/2)mv 2 ] Could we have surmised this result through direct observation (similar to how we first deduced that momentum was P = mv)?

11/13/15 Oregon State University PH 211, Class #215 Transferring Energy by Doing Work If we push on an object of mass m with a steady net force, F net, while that object moves through a displacement,  x, in the direction of the force, how will the object’s properties change? By Newton’s Laws, we know it will accelerate (and at a steady rate, since F net is steady). And we can calculate the resulting motion, using kinematics: (Try with simple numbers: Let m = 3 kg, F net = 15 N,  x = 4 m, and suppose the object starts with v i = 2 m/s.)

11/13/15 Oregon State University PH 211, Class #216 If we do this over and over—with any sort of push through any distance—we notice something peculiar: The quantity (1/2)mv 2 seems to be the property that changes when a net force is applied to an object while the object moves through some displacement. Since (1/2)mv 2 is associated with motion, we call it kinetic energy. Q: What are the units of kinetic energy? Q: And why do we call Joules “energy”? A: Because these same units turn up in all sorts of other phenomena in physics—other ways or abilities to affect other objects. And apparently all those various forms of energy can interchange—but the total number of Joules in the universe doesn’t change. This is the principle of Conservation of Total Energy.

A car travels at some speed, v. If the car’s speed then doubles, by what factor does its kinetic energy increase? A) K will also double. B) K will triple. C) K will quadruple. D) It depends on the mass of the car. E) It depends on the initial speed of the car. 11/13/15 7Oregon State University PH 211, Class #21

11/13/15 Oregon State University PH 211, Class #218 This process of giving an object some more energy—by pushing on it with a net force through some distance—is what we call work. (Notice that the units of work are indeed energy.) Q: Is energy a vector or a scalar quantity? But notice: Work is done on an object only to the extent that the motion is in the same direction as a net force. A given force F may have only a vector component acting as a net force in the direction of the displacement,  x. In other words: Work = F ･  x = Fx(cos  ), where  is the angle between the vectors F and  x.

This “s” is important. This really means: 11/13/15 9Oregon State University PH 211, Class #21

Which force does the most work? 1. The 6 N force. 2. The 8 N force. 3. The 10 N force. 4. They all do the same amount of work. 11/13/15 10Oregon State University PH 211, Class #21

T You pull this 5.0 kg crate along a carpeted floor (  k =.80) with a rope of T = 75N at an angle of 30.0°above the floor. What is the speed of the crate after 1.0 meters of pulling? 11/13/15 11Oregon State University PH 211, Class #21

Example: A 1000 kg car is rolling slowly across a level surface at 1.0 m/s, heading toward a group of small innocent children. The doors are locked, so you can’t get inside to use the brakes. Instead, you run in front of the car and push on the hood at an angle 30° below horizontal. How hard must you push to stop the car in a distance of 2.0 m?  11/13/15 12Oregon State University PH 211, Class #21