PH1110 Term A98
STUDY GUIDE 3: Work, Energy, and Momentum

Objectives

15. Define work and calculate the work done by a constant force as the body on which it acts is moved by a given amount. Be able to calculate the scalar product of two vectors.

16. Define kinetic energy.

17. State the work-energy theorem, Give examples of and solve problems for which the application of the work-energy theorem is appropriate.
18. Define power, and use the concept to solve problems involving the rate at which work is done.

19. Distinguish between conservative and non-conservative forces and give examples of each
.
20. Calculate the change in potential energy of a particle in a uniform gravitational field and of a spring undergoing compression or extension.

21. Use the principle of mechanical energy conservation to solve appropriate problems.

22. Define the linear momentum of a particle and of a system of particles.

23. Define impulse of a force and relate it to the change in linear momentum that it causes.

24. Give examples of and solve problems for which conservation of linear momentum is appropriate. Distinguish between elastic and inelastic collisions.

 Suggested Study Procedure for Chapter 6. Study Secs. 6-1 through 6-5.
Answer Discussion Questions 1, 2, 11, 12, 13.
Study particularly Examples 1, 2, 4, 6, 7, 8.
Do Exercises 1, 3, 13, 15, 19, 23, 27, 30, 37.
Do Problems 47, 57, 65, 71.
  A. As great as we have already found Newton's laws to be in helping us understand and predict the motion of objects, there are wondrous new and enormously powerful concepts lurking in those three disarmingly simple statements. Two such concepts that tumble mathematically right out of the 2nd law are WORK and ENERGY, the subjects of Chapts. 6 and 7. 1. A FORCE carried through a DISPLACEMENT performs WORK. But be careful here! Work is NOT force times distance; rather, it involves only the component of force parallel to the displacement. In the special case of straight-line displacement with constant force (the case we will consider most of the time!), work is equal to the product of the force component parallel to the displacement with the distance traveled.

2. Work and energy are SCALAR quantities -- they are purely numbers WITHOUT spatial direction. DO NOT attach vector directions to work or energy; that's just plain WRONG! Now it may bother you that a scalar can result from the multiplication of two vector quantities (force and displacement), but that's the way it is. The mathematical operation that accomplishes this strange feat is called the "scalar product" and is discussed in detail in Sec. 1-11. Secs. 6-1 and 6-2 provide discussion and worked examples on the subject.
 

 B. The WORK-ENERGY THEOREM (which comes mathematically straight out of the 2nd law!) states that the TOTAL WORK done on an object is equal to the CHANGE IN KINETIC ENERGY of the object (final minus initial). What's so great about that, you may ask? Well, for one thing, a scalar equation is invariably easier to work with than a vector equation. For another, we're concerned with speeds of the object only at the beginning and at the end of the displacement, and not anywhere in between. If you know an object's speed at the beginning and end of the object's displacement, you immediately know how much total work was done on the object. If you know the total work done, you immediately know the change in kinetic energy. The power of this generalization in analyzing motion is demonstrated repeatedly from Example 6-4 on through most of the exercises and problems at the end of the chapter.

C. The one non-constant force we consider in PH 1110 is the spring force, discussed in Sec. 6-4. Study this section carefully to see how work is properly computed when the force varies linearly with displacement.

D. POWER, the RATE at which WORK is performed, is also a scalar quantity. Read about it in Sec. 6-5.

 Suggested Study Procedure for Chapter 7. Study Secs. 7-1 through 7.4.
Answer Discussion Questions 1, 3, 4.
Study Examples 1, 2, 5, 6, 7, 8, 9, 10, 12.
Do Exercises 3, 7, 13, 15, 17, 25.
Do Problems 40, 41, 53, 63, 65.
  A. Believe it or not, Chapt. 7 is just one long series of applications of the Work-Energy Theorem, even though it may not look like it! As you found in Chapt. 6, the TOTAL work is just the (scalar!) SUM of the work done by each individual force. Because the work done by CONSERVATIVE FORCES (the gravitational and spring forces) depends only on the endpoints of the displacement (NOT how we got from the initial to final point!), we can pull them out of the TOTAL WORK and express them in terms of initial and final points. The way we like to express the theory of Chapt. 7 is:
  ½mvi2 + mghi + ½ksi2 + Wother = ½mvf2 + mghf + ½ksf2
  This is what Chap. 7 is all about, and Lecture #10 will be aimed at making its application crystal clear!

Suggested Study Procedure for Chapter 8.

Study Secs. 8-1 through 8-5.
Answer Discussion Questions 6, 11, 12.
Study Examples 2, 3, 4, 5, 6, 7, 8, 9, 10.
Do Exercises 1, 3, 8, 14, 23, 26, 35.
Do Problems 60, 62, 64, 65, 73, 74.
  A. Two more extremely powerful concepts that tumble mathematically out of Newton's 2nd law are IMPULSE and MOMENTUM. IMPULSE is a FORCE acting through TIME -- a sort of product of force and time, with units of newton-seconds. MOMENTUM is the product of mass with velocity. According to the 2nd law, the IMPULSE received by an object equals the CHANGE OF MOMENTUM of the object. 1. Note the similarity with the form of the Work-Energy Theorem (where TOTAL WORK equals the CHANGE IN KINETIC ENERGY). Also note that, just as in the case of work-energy, we are concerned with a change from an initial to a final state, without worrying about precisely what happened in between. That's the real power of this IMPULSE-MOMENTUM stuff, just as in the case of WORK-ENERGY.

2. But PLEASE also note the PROFOUND DIFFERENCES! For example, impulse and momentum are VECTOR quantities, meaning that direction is an integral attribute of each, and problem solutions will require use of components. Nonetheless, this complication is more than offset by the power of the approach, as illustrated in Sec. 8-1 and 8-2.

 B. Impulse-Momentum is especially important in considering the COLLISION of two or more objects. Here's why! Remember Newton's 3rd law? All those action-reaction pairs of forces between interacting objects are equal and opposite. Thus, if you add up all the impulses resulting from those action-reaction pairs, the sum must be ZERO! That is, the NET IMPULSE must be ZERO. Zero net impulse requires that the TOTAL MOMENTUM of the system of two or more objects NOT CHANGE during the collision. In such cases, we say that momentum is a CONSERVED QUANTITY (a constant). Secs. 8-3 through 8-5 show how this important principle of CONSERVATION OF MOMENTUM is applied to the solution of collision problems.

C. When two or more objects collide in the absence of net external force, the total momentum of the system of objects is ALWAYS AND NECESSARILY CONSERVED. The total kinetic energy of that system of objects, however, is seldom the same afterwards as it was before. In those RARE situations where kinetic energy is conserved, we call the collision "elastic." NEVER assume that a collision is elastic UNLESS you are told that it is OR you calculate that it is after you have solved the problem using momentum conservation. All other collisions where the kinetic energy changes during the collision are called "inelastic." Read all about it in Secs. 8-4 and 8-5.

 HOMEWORK ASSIGNMENTS FOR STUDY GUIDE 3.

Homework Assignment #8 - due in lecture Monday, Sept 21.

Yes, we know that we have not yet touched on this material in lecture, but these are really straightforward exercises, so you should be able to dig out the necessary theory on your own!

Exercise 6-2, except change the mass of the crate to 50.0 kg and the coefficient of friction to 0.250. Also, draw a free-body diagram for the crate AND draw a sketch of the situation, including a coordinate system, summarizing the key details at issue here. THIS IS SOMETHING THAT YOU SHOULD CONTINUE TO DO WITH ALL PROBLEMS THROUGH TO THE END OF THE COURSE!

Exercise 6-6, except the bucket and contents have a total mass of 12.0 kg and are raised a total of 5.00m. (Remember to draw a free-body diagram)

Exercise 6-8, except the vehicle is a 2500 kg truck travelling at 50.0 km/hr. Also, compare the kinetic energy in (a) to that of a 5000 kg vehicle traveling at 25.0 km/h.

 Homework Assignment #9 - due in lecture Wednesday, Sept 23. Exercise 6-48, except that a 5.00 kg package slides 4.00m down a long ramp that is inclined 20.0° below the horizontal and has a coefficient of friction of 0.400. (Remember to draw a free-body diagram!).

Problem 6-64.

Problem 6-70; Draw free-body diagrams for both blocks and compute the work done by all of the forces present in the system. (Note that you don't need to know the actual value of the rope tension force to do this calculation! Why not?). Once you have evaluated the change of speed in this problem, calculate the numerical value of the tension force in the rope.

 Homework Assignment #10 - due in lecture Friday, Sept 25. Problem 7-47, except an 8.00 kg mass is substituted for the 4.00 kg mass. Once you solve for the speed after the 2.00m travel, solve for the tension force in the rope and the time required for the 2.00m motion.   Problem 7-48, except the skier snowplows down the hill, arriving at the bottom with only 70% of the speed of a frictionless trip, then encounters 225m of m k = 0.12 rough snow, and finally plows 3.00m into the snowdrift. Also, solve for (d) the energy dissipated by snowplowing down the hill, and (e) the ratio of the average snowdrift force to the skiers weight (i.e., the g-force) during the final slowdown.

SNJ #6: Similar situation to Problem 7-40, except the incline has a coefficient of friction of 0.25 (the horizontal surface is still frictionless) and the block is released from rest at a point 3.00m up along the incline.
      1. Calculate the speed the block will have when it reaches the bottom of the incline and slides toward the spring.
      2. Calculate the distance the spring compresses in bringing the block to rest.
      3. Calculate the distance the block travels back up the incline on its first rebound.
 

Homework Assignment #11 - due in lecture Monday, Sept 28.

Exercise 8-10; Before solving for the average force on the ball, calculate the impulse received by the ball from the bat and the impulse received by the bat from the ball.

Exercise 8-32; Also, calculate the impulse received by each car during the collision.

SNJ #7: Similar situation to the one described in Problem 8-60, except the tennis ball is hit so as to fly back along its incoming path with the same speed. Solve for the impulse received by the ball. Given that the ball is in contact with the racket for 4.00ms, solve for the average force exerted by the racket on the ball (both in component form and as magnitude/direction).

 Homework Assignment #12 - due in lecture Wednesday, Sept 30. Problem 8-72.   SNJ #8: Similar situation to Exercise 8-23, except that Puck B has twice the mass of Puck A: mB = 2mA = 0.320 kg. The incoming velocity and outgoing angles remain the same. Do (a) and (b) as stated in 8-23. Also solve for the impulse received by each puck, the sum of these impulses, and the average force acting on each puck, if the contact time is 3.00ms..

SNJ #9: Similar geometry to Problem 8-64, but mA = 4.00 kg, m B = 1.00 kg, m C = 5.00 kg, all three objects have a speed of 2.00 m/s prior to colliding at the origin, and the velocity of object C is the negative of object B's velocity.
    1. Given that the three objects stick together after colliding, calculate the final velocity of the three-particle clump.
    2. Calculate the percentage change of the total kinetic energy and explain whether it is a gain or a loss.