Work W= Fd W TOT = Σ Fd Work – energy theorem W TOT = Δ KE + Δ PE

2. Work • W=Fd • WTOT=ΣFd • Work – energy theorem • W TOT= ΔKE + ΔPE

3. Law of conservation of mechanical energy • KE + PE = constant • Mechanical Energy = KE + PE

4. Kinetic energy • KE = ½ mv2 • ΔKE = ½ mvf2 - ½ mvi2 • Potential energy • PE = mgh • ΔPE = mgh1 – mgh2 • Momentum • p = mv • Δp = mΔv • Δp = Δmv • Impulse • j = Ft

5. Law of conservation of momentum • pbefore = pafter • mavai+mbvbi= mavaf+mbvbf • Impulse-momentum theorem • Ft = mΔv

7. Work All effort without affecting motion is no work at all. Effort is only given credit if it results to something sensible. There is no hard work if there is no result. You are paid for your work if your effort results to the direction expected of you.

9. A 10 N box is pushed from rest by a 20N force toward east against a 10 N friction for a distance of 10 m. • A 10.0 kg crate experiences a 10.0 N force east and a 10.0 N force west as it moves a distance of 10.0 m to the east at a speed of 2.00 m/s. • A 10 N object experiences two forces F1 = 10 N west and F2 = 20 N east. These two forces are applied at the same time until the object covers a distance of 10 m in the direction of F1. • An object is pulled up by a 10 N force at constant velocity to a height of 10 m. • A 10 N object is brought down from a 10 m shelf to the ground at constant velocity. • A 10 N object slides on a frictionless floor a distance of 10 m as it is pulled by a 10 N force that is oriented 30o with the horizontal.

10. For each situation • Inspect forces and its resulting work • Quantify work and total work on the object • Determine the state of motion based on the magnitude of work. • Determine the change in mechanical energy of the object. • Quantify the kinetic energy or/and potential energy of the object • Quantify the momentum and change in momentum of the object

11. Target Skill • Ability to qualify situations for the presence of work and the forces responsible for it • Ability to quantify work, power, momentum, impulse, kinetic energy and potential energy • Ability to inspect changes in momentum • Ability to inspect changes in kinetic energy • Ability to inspect changes in potential energy

12. Skill Check Point • An 8.00 kg object starting from rest is pushed by a 10.0 N force against a 5.00 N friction for a distance of 1.00 m. • (3) Does the 10 N force perform work on the object? Why? • (3) Does friction perform work on the object? Why? • Find: • (5) The work performed by the given forces • (5) The magnitude of total work on the object • (7) The change in the kinetic energy of the object • (8) The change in the potential energy of the object • (6) The change in the momentum of the object • (5) The momentum of the object when it is halfway from its starting point

13. Answers • Yes, the 10 N force performs work on the object because it directly affects the resulting displacement. • Yes, friction performs work on the object because it directly affects the resulting displacement by opposing the propelling force. • The work performed by the 10 N force is 10.0 J while the work performed by friction is – 5.00 J. • The total work on the object is 5.00 J resulting from the work performed of F and f. • The change in kinetic energy is 5.00 J, since the change in kinetic energy is just equal to the total work. • The object does not change in potential energy because it does not move in the y-axis. There is only change in potential energy if the object changes its height from the ground.

14. Answers • The initial momentum is zero. To find the change in the momentum of the object, the final velocity is required so that we can find the final momentum. The change in momentum is simply momentum final minus momentum initial (Δp=mvf - mvi). We can find vf in two ways: • Finding acceleration from the second law and using kinematics (demonstrated yesterday). • Using the equivalence equation of total work and change in kinetic energy.

15. The equivalence of total work and a change in kinetic energy gives: • Since initial velocity is zero, then the equation becomes • Now that the final velocity is known, the change in momentum is determined by Δp=mvf because the initial momentum is zero. Therefore the change in momentum is 8.96 kg.m/s.

16. The momentum halfway from its starting point is the momentum of the object when it is 0.50 m away from its origin. To find this momentum which is simply p=mv, determine first the velocity at that point. • The acceleration of the object is 0.625 m/s2 determined by using the Law of Acceleration. • The acceleration, displacement and initial velocity are known. Therefore the velocity at 0.500 m from starting point is 0.791 m/s from

17. The acceleration, displacement and initial velocity are known. Therefore the velocity at 0.500 m from starting point is 0.791 m/s from • Therefore, the momentum of the object when it is halfway from its starting point is 6.33 kg.m/s • YAY! 

18. Lifting an Object • The energy you utilized in lifting an object (which is work) is transferred to the object in the form of gravitational PE and KE.

19. Throwing a javelin • The energy utilized in throwing the javelin (which is another form of work) is transferred to the javelin in the form of KE and PE.

20. Baseball Pitch • Throwing a baseball transfers energy to the ball.