Chapter 6 Work and Energy

Chapter 6 Work and Energy

Midterm exams will be available next Thursday.

Assignment 6 Textbook (Giancoli, 6 th edition), Chapter 6: Due on Thursday, November 5 1. On page 162 of Giancoli, problem 4. 2. On page 162 of Giancoli, problem 8. 3. On page 163 of Giancoli, problem 24. 4. A running man has half the kinetic energy that a boy of half his mass has. The man speeds up by 1.00 m/s and then has the same kinetic energy as the boy. What were the original speeds of man and boy?

Chapter 6 Work Done by a Constant Force Kinetic Energy, and the Work-Energy Principle Potential Energy Conservative and Nonconservative Forces Mechanical Energy and Its Conservation.

Recalling Last Lecture

Kinetic Energy, and the Work Energy Principle In mechanics, we can use a less precise definition of energy as: The ability to do work The energy of motion is called kinetic energy (or KE for short).

Kinetic Energy, and the Work Energy Principle Work-energy principle: The net work done on a object is equal to the change in the object s kinetic energy 1) If W net > 0, then KE > 0 implying that there was an increase in the object s velocity ( v 2 > v 1 ). 2) If W net < 0, then KE < 0 implying that there was an decrease in the object s velocity ( v 2 < v 1 ).

Kinetic Energy, and the Work Energy Principle Problem 6-25 (textbook): A 285-kg load is lifted 22.0 m vertically with an acceleration a = 0.160g by a single cable. Determine (a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the load, (d) the work done by gravity on the load, and (e) the final speed of the load assuming it started from rest.

Problem 6-25: (a) Kinetic Energy, and the Work Energy Principle From the free-body diagram for the load being lifted, write Newton s 2 nd law for the vertical direction, with up being positive. F net = F T F = F m g = m a = 0.160 m g T ( )( ) = = = 2 3 1.16m g 1.16 285 kg 9.80 m s 3.24 10 N F r T m g r

Problem 6-25: (b) Kinetic Energy, and the Work Energy Principle The net work done on the load is found from the net force. W = F d = mg d = net net = ( ) ( )( )( ) o 2 cos 0 0.160 0.160 285 kg 9.80 m s 22.0 m 3 9.83 10 J F r T m g r

Problem 6-25: (c) Kinetic Energy, and the Work Energy Principle The work done by the cable on the load is W F d mg d ( ) ( )( )( ) o 2 4 = cos 0 = 1.160 = 1.16 285 kg 9.80 m s 22.0 m = 7.13 10 J cable T F r T m g r

Problem 6-25: (d) Kinetic Energy, and the Work Energy Principle The work done by gravity on the load is W mgd mgd ( )( )( ) o 2 4 = cos180 = = 285 kg 9.80 m s 22.0 m = 6.14 10 J G F r T m g r

Work Done by a Constant Force Problem 6-25: (e) Use the work-energy theory to find the final speed, with an initial speed of 0. W = K E K E = m v m v v 1 2 1 2 n e t 2 1 2 2 2 1 n e t 2 2 1 3 ( ) 2W 2 9.8 3 1 0 J = + v = + 0 = 8.3 1 m s m 2 8 5 k g F r T m g r

Potential Energy Another form of energy carried by an object is called potential energy. Potential energy is associated to forces that depend on the position or configuration of an object. There are several forces that depend on either the configuration or position of an object therefore there are several different kinds of potential energies, each associated to a different force. For example, we have seen that the gravitational force depends on the distance between two objects therefore, the energy of an object associated to the gravitational force (gravitational potential energy) will consequently depend on the position of the object. Let s analyze the gravitational potential energy..

Potential Energy In raising a mass m to a height h (no net acceleration), the work done by the external force is (6-8) Gravity also does work on the object of mass m: Now, note that if at any instant of time you drop the object, it can do work on something located underneath it. Let s say it strikes a nail, moving it into the floor. it will change the nail s velocity, therefore it will change the nail s kinetic energy.

Potential Energy How does it work? As you drop the object, it will immediately gain velocity due to the acceleration of gravity. Assuming its initial velocity to be zero, we have for its velocity when it strikes the nail (travels a distance h): Therefore, using the work-energy principle, we have: 0 2gh This results tells us that an object located at a height h has the ability to do an amount of work given by mgh

Potential Energy The previous result can be generalized to any height y such that we can define a gravitational potential energy as (6-9) Given that h = y 2 y 1, it follows that: which is exactly eq. 6.8 obtained two slides ago. We can rewrite the above equation as: (6-10) Note 1: The choice of y=0 is arbitrary only changes in potential energy matter. Note 2: Again, from two slides ago we can conclude that:

Potential Energy Important: 1) Potential energy is a property of a system as a whole. For example, when you lift the object in the figure from y 1 to y 2, the work done is given by the difference between PE 2 and PE 1. The values of PE 1 and PE 2 depend on the object and on the Earth (responsible for the force of gravity), therefore on the complete system: Earth + object. This is NOT the case with kinetic energy where ONLY the properties of the object matters 2) The work done by the gravitational force on an object depends only on the height and not on the path taken by the object.

Potential Energy (Elastic force) As mentioned before, there are other forms of potential energy. I will now discuss on one of them associated with elastic materials. A spring has potential energy when compressed or stretched. For example, the figure below shows the potential energy stored in a spring when it is compressed yielding kinetic energy when it is released.

Potential Energy (Elastic force) The force needed to compress or stretch a string is written as: Where, k = spring stiffness constant Newton s 3 rd law says that the spring will exert a force F S in reaction to the force F P such that: Textbooks usually write: (6-11) Eq. 6.11 (spring equation) is known as Hooke s law F S is the force the spring exerts opposite to its displacement in an attempt to restore its original length

Potential Energy (Elastic force) Now that we know the force, we can define the potential energy of a spring compressed or stretched by a length x from an initial reference position. The work done to change the spring s natural length, let s say stretch it, will be given by: Where, (We cannot use F P directly to calculate the work done on the string since F P = kx varies with the position x. Thus, we use the average force applied on the spring). Thus We can define the elastic potential energy as: (6-12)

Kinetic Energy, and the Work Energy Principle Problem 6-30 (textbook): A 1.60-m tall person lifts a 2.10-kg book from the ground so it is 2.20 m above the ground. What is the potential energy of the book relative to (a) the ground (b) and the top of the person s head? (c) How is the work done by the person related to the answers in parts (a) and (b)?

Kinetic Energy, and the Work Energy Principle Problem 6-30: (a) Relative to the ground, the PE is given by 2 ( ) ( )( )( ) PE = mg y y = 2.10 kg 9.80 m s 2.20 m = 45.3 J G book ground (b) Relative to the top of the person s head, the PE is given by 2 ( ) ( )( )( ) PE = mg y y h = 2.10 kg 9.80 m s 0.60 m = 12 J G book head (c) The work done by the person in lifting the book from the ground to the final height is the same as the answer to part (a), 45.3 J. In part (a), the PE is calculated relative to the starting location of the application of the force on the book. The work done by the person is not related to the answer to part (b).

Kinetic Energy, and the Work Energy Principle Problem 6-29 (textbook): A 1200-kg car rolling on a horizontal surface has speed v = 65 Km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?

Kinetic Energy, and the Work Energy Principle Problem 6-29: Assume that all of the kinetic energy of the car becomes PE of the compressed spring. mv ( 1200 kg) ( 65 km h) 2 1 2 1 2 4 mv = kx k = = = 2 2 2 2 x ( 2.2 m) 1m s 3.6 km h 2 8.1 10 N m