Work, energy, power, and conservation of energy

Transcription

1 Work, energy, power, and conservation of energy We ve seen already that vectors can be added and subtracted. There are also two useful ways vectors can be multiplied. The first of these is called the vector dot product, or just dot product. Suppose we wanted to take the vector dot product of A and C. It would be written A C. The heavy dot between the two indicated the dot product. It s very easy to compute. A C = A x C x + A y C y = (4 1) + (2-4) = -4. The answer is just a number without direction, which is a scalar, so the dot product can also be called the scalar product. All you are doing is multiplying the two x-components, multiplying the two y-components, and then adding those two products together. Try it with A B and see what you get. A B = A x B x + A y B y = (4 2) + (2-4) = 0 Conceptually, the dot product is a measure of how parallel two vectors are to each other. Vectors A and B are perpendicular, so they are not at all parallel, making the dot product zero. You can also see fairly easily that A C = C A because A x C x + A y C y = C x A x + C y A y If you take the dot product D E or E D, you will get a high negative number, -20, because the two vectors are highly anti-parallel.

2 Another way of computing the dot product, useful in other contexts, is to use the cosine function. Let s use A C again. Here, A C = A C cosθ This is read as A-dot-C equals the magnitude of A times the magnitude of C times cosine of the angle between the two vectors. The magnitude of A we can get from the Pythagorean Theorem, A = = 20 Likewise, C = 17 Now, what is the angle between the two vectors? Let me draw them with a common origin. If you use the inverse tangent function, you will find A has an angle of º and C has an angle of 284º. The angle used in the dot product is the angle from the first vector, counter-clockwise, to the second vector. Because we are working with A C, we want the angle swept from A to C. This angle is º, making the full dot product: A C = cos(257.47º) = -4. This is the same result as before. It is true that if you swept out the angle clockwise, you would get the same answer. It doesn t matter for the dot product because the cosine function is symmetric about the y-axis. But it will matter later for the vector cross product, so the counter-clockwise sweep is a good habit to begin. Answer Webassign Question 1 How these two definitions of the dot product are the same, I will leave for an appendix at the end.

3 In physics, the concept of work is defined as the dot product of force and displacement. W = F Δs or, using the second method described above, W = F Δs cosθ Work is the magnitude of the force times the magnitude of the displacement times the cosine of the angle between them. Again, any dot product is a scalar, so the physical concept of work is a scalar. If the force is not constant over the displacement, calculus notation is required. dw = F ds a little work is input by a force over a little displacement dw = F ds W = F ds Let s see how this can be used in an example. Suppose you push a 20kg crate up a 30º frictionless incline by applying a force of 150N directly up the incline over a distance of 4m. Now, I ll redraw the situation with all of the relevant vectors and then a vector diagram off to the side. Because there are three forces during the displacement, there are three works to calculate: W P = F P Δs cosθ W G = F G Δs cosθ W N = F N Δs cosθ

4 Although they all have a common displacement, the forces and angles will be different for each. The force of the push is given as 150N, the force of gravity is 196N (again, magnitudes only), and the normal force on an incline is m g cos(θ incline ) = 170N For each, the angle is from the force vector, counter-clockwise, to the second vector. For the push, the angle between the push force and the displacement is 0º. For the force of gravity, the angle from the force of gravity to the displacement vector is 240º. And for the normal force, this angle is 90º. Putting all six of those numbers into the equations above, with a displacement of 4m yields: W P = 600N m W G = -392N m W N = 0N m Because work is a scalar, we can simply add up all of these numbers to calculate the net work: ΣW = 208N m This can also be written as 208J, as 1 Joule = 1 Newton-meter. Answer Webassign Question 2 The concept of potential energy can be related to work with the idea that the work done by a conservative force equals the negative change in potential energy: W conservative = -ΔU What is a conservative force? One way of defining it is to say that the work it does is independent of the path taken by an object. This may not be very helpful, but let me just say that the two types of forces initially, spring forces and gravitational forces, are both conservative forces. So there can be spring potential energy and gravitational potential energy.

5 Let s look at gravitational potential energy first. Suppose you take a stone with mass m and lift it from the ground to a final height s y. What is the gravitational potential energy of the stone? First, calculate the work done by the force of gravity. W G = F G Δs cosθ = (m g)(s y )cos180º = -m g s y This is equal to the negative change in potential energy, W conservative = -ΔU -m g s y = -ΔU or ΔU = m g s y s y is more commonly known as height, h, so ΔU = m g h If we declare that the potential energy on the ground is zero, then the final potential energy is the same as the change and you end with an energy, U gravitational = m g h. Technically, this potential energy is not held in the stone, but rather in the separation between the Earth and the stone. Answer Webassign Question 3 Suppose you had instead lifted the same stone to the same height along a curved path: Would the work done by gravity be any different?

6 You can see that it won t be if you imaging zooming-in on a small section of the curve. In the light-blue horizontal steps, no work is being done by gravity because the force of gravity and those little displacements are all perpendicular. But when you add up all of the dark blue vertical steps, you get the same displacement as a direct, vertical lift. So the work done is the same as if it was simply lifted straight upwards. This is the meaning of gravity being a path independent force. Now suppose you have a box attached to a spring, which begins unstretched at position zero. You then pull the box to a new position, s x. How much energy will the spring store? Again, we need to first find the work done by the spring, W spring = F spring Δs cosθ But the spring is not going to apply a constant force; the further it is stretched, the stronger will be the force it applies. In fact, the graph of spring force versus position likely looks like this:

7 which follows the equation, F spring = -k x, where k is the spring constant of the spring and k is the slope of the line. What we can do is find the area of the force-position graph and take this as the work done by the spring. This area is k x2 and so W spring = k x2 If W = -ΔU, then ΔU spring = 1 2 k x2 Again, declaring the initial potential energy of the spring to be zero, U spring = 1 2 k x2 Answer Webassign Question 4 For a conservative for, the equation W = -ΔU can be written in differential form as dw = -du. If we then write dw = F ds for work in one-dimension, F = - du This equation is useful when analyze graphs of potential energy versus position. ds Aside from generating potential energy, work can also generate what is known as kinetic energy. For example, suppose a block with a mass of m sits at rest on a frictionless surface. If someone pushes the block with a force of F over a displacement Δs, they will input work to the block. W person = F Δs = (m a) Δs from Newton s second law of motion and if v 2 f = v 2 i + 2(a)(Δs), then (a)(δs) = v f 2 when the initial velocity is zero 2

8 W person = m vf 2 2 = 1 2 m v f 2 So here the work input is producing a quantity of motion known as kinetic energy, K = 1 2 mv2. Answer Webassign Question 5 Related to work is the concept of power. In the same situation of someone pushing a block along a frictionless surface with a force and changing the block s kinetic energy, there was a certain amount of power input to the block by the person. In that example, we could calculate the power input as the total work done by the person divided by the change in time in which it was input. P = W Δt Generally, work is a way of transferring energy from one system to another through the application of a force. There are many other ways of transferring energy, heat, for example. So a more general definition of power is energy transferred per time. P = ΔE Δt Also, if one replaces the work in the first equation for power with F Δs, it becomes P = F Δs Δt = F v Power at some instant in time is the force applied to an object times the velocity of that object. The equation we will use for conservation of energy is this: ΔE = W + Q ΔE is the total change in energy of the system, which could be in the form of thermal energy, chemical energy, gravitational potential energy, spring potential energy, and kinetic energy. There are other ways to categorize energies, but these are the ones we will use. W is the work input to the system Q is heat flowing into the system (+) or out of the system (-). This is useful in thermodynamic processes like engines and refrigerators but which are not part of this class, so Q will most likely always be zero.

9 So let s see how this can be applied to a certain situation. A cart begins at the top of a plateau, pushed into a spring a certain distance. The cart begins at rest, but when the spring is released, the cart rolls down the incline and along the floor. How this situation is analyzed in terms of work and energy depends upon what is chosen as the system and what is chosen as the environment. Let s start by just taking the cart itself as the system. Does the system begin with kinetic energy? No, nothing is moving at first. Does it begin with gravitational potential energy? No. Even though the cart is above the ground, we have chosen the cart alone as the system and there is no cart/earth separation in this system. Does it begin with spring potential energy? No. Again, the compressed spring is not part of the system as defined. What does work on the system? The spring does work because it is a force applied to the system from outside the system which occurs over a displacement of the cart. Likewise for the force of gravity as the cart descends the ramp. Lastly, the cart ends with kinetic energy as it rolls along the floor. So ΔE = W + Q would be written as 1 2 mv2 = W spring + W gravity

10 Let s use the same situation, but define the system as the cart and spring together. To begin: K = 0 U gravitational = 0 U spring = 1 2 kx2 When the spring is fired, however, it does no work on the system because it is internal to the system. In the firing, there is a change in spring potential energy of kx2. In the descent, gravity does work on the cart. Altogether, we have: kx mv2 = W gravity Suppose you focus on everything, including the Earth. Now we do begin with spring potential energy and gravitational potential energy, because the compressed spring and Earth/cart separation are included in our system. The spring force and gravitational force are internal to the system, so they input no work. Altogether, we have kx2 + (-mgh) mv2 = 0 There s one further technical point to be made. Suppose we take as our system a block sitting on the floor at rest. When the block is treated as an object, we can say that it has zero energy. Conceptualizing the block as an object assumes it has no internal structure.

11 But if we zoomed in to the particles which make-up the block, we are then conceptualizing the block as a system of particles. Now the block does have an internal structure and it can have kinetic energies of the particles vibrating in place and potential energies within the molecular and atomic bonds. This sum of kinetic and potential energies is the internal energy of the block. When a situation like the one described above is analyzed in such a way that the object is the system, there is no possibility of potential and internal energies, so the conservation of energy equation reduces to ΔK = ΣW, sometimes called the work-energy theorem. At the other extreme, one can take the entire universe to be the system, so that external work and heat flowing into the system are impossible and the conservation of energy equation reduces to ΔE = 0. To review: A B = A x B x + A y B y = A B cosθ W = F Δs W = F ds if the force is constant over the displacement if it is not The area under a force/position function is the work input F = - du ds K = 1 2 m v2 U gravitational = m g h and U spring = 1 2 k x2 P = ΔE Δt or P = W Δt = F v ΔE = W + Q

12 Appendix proof of the equivalence of the dot product definitions If you look at the vector diagram above, A + (B A) = B is a valid relationship The law of cosines states c 2 = a 2 + b a c cosθ Here, that would be: (B A) 2 = A 2 + B A B cosθ Expanding the left side yields B 2 2A B + A 2 = A 2 + B A B cosθ Something like A A = A A cos(0º) = A 2, therefore B 2 2A B + A 2 = A 2 + B A B cosθ So A B = A B cosθ