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Transcript
CHAPTER 5 WORK AND ENERGY So far we have been looking at problems and analysing them using Newton’s laws of motion. We now consider another method of solving problems. One which we will find useful when dealing with objects moving at speeds close to that of light or for particles that are very small (like the electrons and protons that make up atoms). In such cases Newton’s laws of motion need to be modified before they are valid. Our new methods are always valid. They invlove terms that you are already familiar with but as always physicists give them very precise meanings. Work: motion in 1-dimension with a constant force Let us assume that a student pushes an object along a frictionless surface. The force he applies F is parallel to the surface and the resulting straight line motion. He pushes the object through a displacement d. We define a quantity work, W, as given by W = Fd where F is the magnitude of F and d the magnitude of the displacement d. We say that W is the work done by F in moving the object through a displacement d. The work is done by a force. Note that work is a scalar quantity although force and displacement are vectors. F F d d Now what if the force was not parallel to the motion, as shown in the figure below. In this case we define the work to be W = Fd cos where is the angle between the force and the displacement vectors (when placed tail to tail). We can think of it as the component of the displacement in the direction of the force or the other way around (they are equivalent). Notice that for our initial situation = 0 so our more general equation reduces to the special case. F F d 31 d Sometimes physicists like to write the above equation in a more compact form using the scalar (or dot) product. It is another way of saying include the cosine of the angle between the 2 vectors W = F.d Our more general equation for work has some more special cases. If the force and displacement vectors are in opposite directions then = 180 so that cos = -1. Since F and d are always positive this means the work done in such a situation will be negative. If the force and displacement vectors are perpendicular to each other then = 90 so that cos = 0. In this case there is no work done. To illustrate these possibilities consider the following situation. You pick up a chair, carry it with a constant velocity across the room and place it back down again. In the first part you apply a force F to lift the chair. The force is upwards and so is the displacement. The force (you) does positive work on the chair. While you are walking across the room with constant velocity holding the chair you still apply an upward force F on the chair (otherwise it would fall) but the displacement is perpendicular to this. The force (you) does no work in this case. Finally as you place the chair down you still exert an upwards force (as you place the chair down slowly) but the displacement is downwards so the force does negative work. Note that throughout the lifting and carrying you must exert a force but that does not mean you are always doing work, at least not as far as physics is concerned. Your muscles are straining to support the weight of the chair and they are consuming energy but not necessarily doing work. The SI unit for work is the newton-metre since they are the units for force and displacement. However, work is such a useful quantity that the unit has been given a special name, the joule (J). It is the work done by a force of 1 N moving through a displacement of 1 m. As we have previously seen there can be several forces acting on an object. How do we calculate the work done in such a case? There are 2 ways. We can first calculate the net force acting on the object and then use our general equation for work. Or we can calculate the work done by each of the forces and simply add them up. Either way we obtain the same answer. Q. A weight lifter raises a mass of 250 kg a distance of 2 m. How much work has he done? A. Drawing a free-body diagram To find the force exerted by the weight lifter we assume the weight moves at a constant velocity so the net force must be zero. F - mg = 0 F = mg 32 d F The work done is W = Fd cos = (250 kg) (9.8 m.s-2) (2 m) cos 0 = 4900 J. weights Work done by a variable force. mg We now look at the work done by a variable force (one that is not constant). We will assume that the direction of the force is the same as the displacement (i.e. = 0) and then generalise. The force F varies in some arbitrary way that depends on x, F(x), as indicated in the figure below. We cannot calculate the work done using our previous equation because the force is not constant. Instead we divide the motion up into smaller sections, x1, x2, ... xn, each of displacement x. During each of these sections the force does not change much and we approximate it as constant F ( x ) . The work done during each small section is W where W = F ( x ) x F(x) F(x) x xi xf x xi x xf The total work done is approximately the sum of the work done in each of the smaller sections. W = W = F ( x ) x We can make this approximation better by making the sections smaller and taking more of them. In the limit we allow the displacement of the sections to approach zero, x0 and we obtain the work done by the variable force F(x). W = lim x 0 F ( x ) x This limit is what we define as the integral under the curve F(x). So the work done by a variable force is given by W = xf xi F ( x ) dx where xi and xf are the initial and final positions of the object. 33 Variable force in 3-dimensions For an object acted upon by a force F with components F = Fx x + Fy y + Fz z that moves from an initial position (xi, yi, zi) to a final position (xf, yf, zf) the work done is given by W = xf xi Fx dx + yf yi Fy dy + zf zi Fz dz This can also be written in the compact notation as W = F dr Work done by a spring As a practical example of a situation where the force on an object varies (and in fact with position) we will look at the force exerted by a spring. A spring that is not being stretched or extended is said to be in its relaxed state. Let us fix one end of the spring (attach it to a wall or hook) and place an object on the other (free) end. To measure the displacement of the object we define our origin to be the position when the spring is in the relaxed state. Furthermore, we consider only 1-dimension, x, along the axis of the spring. When we stretch the spring the there is a force in the opposite direction to the displacement trying to restore the spring to its relaxed state. When we compress the spring the force is again opposite to the displacement trying to restore the spring to the relaxed state. This force due to the spring is called the restoring force. We usually define the x-axis such that when the spring is extended x is positive and when the spring is compressed x is negative. x=0 F=0 x negative F positive x positive F negative F F x x x 0 0 0 Now it turns out that to a good approximation for most springs the force F is proportional to the displacement x from the relaxed state. We write the force as F = -kx and this is called Hooke’s law. The constant k is termed the spring constant and is a property of each individual spring. It is a measure of how stiff (difficult to stretch) the spring is. The SI units for the spring constant are newton per metre (N.m-1). The minus sign in Hooke’s law indicates that the force is in the opposite direction to the 34 displacement. Although the above is a vector equation you will often see it written in terms of the components as F = -kx Hooke’s law gives us the explicit dependence of the force on the displacement so that it is possible to calculate the work done by the spring using our integral method. Let us assume that we move the above object (attached to a spring) from an initial position xi to a final position xf. The work done by the spring is W = xf xi F ( x ) dx = = (½k) x xf xi xf kx dx = k x dx xi 2 xf xi = (½k)( x xi2 ) 2 f Note that since F(x) is the force exerted by the spring it is the work done by the spring. The work done by us is the negative of this (we exert an opposite force). If the initial position of the object (on the spring) is at xi = 0 then the work done in moving it to a final position of xf is W = ½k x 2f Q. An object is attached to a spring with spring constant 500 N.m-1. If the spring is stretched by 10 cm from the relaxed state how much work does the spring do? If it is stretched another 10 cm how much work is done? A. Work done by the spring is given by W = (½k)( x 2f xi2 ) where xi = 0 m and xf = 10 cm = 0.10 m. Substituting we obtain W = (½ (500 N.m-1)) ( (0.10 m)2 (0 m)2 ) = 2.5 J To move a further 10 cm we have xi = 10 cm = 0.10 m and xf = 20 cm = 0.20 m W = (½ (500 N.m-1)) ( (0.20 m)2 (0.10 m)2 ) = 7.5 J {The spring does negative work in each case but the force stretching the spring does positive work.} 35 Kinetic energy We have looked at calculating the work done by a force acting on an object but what about the object. Consider the following. A boy picks up a ball and then throws it. We neglect the initial lifting and focus on the actual throw. During the throwing motion the boy exerts a force on the ball. This force is in the same direction as the motion so it does positive work. When the ball eventually leaves the boy’s hand it is moving with a speed v. What is it about the state of the ball, after the throw, that tells you work has been done on it? There is a property of any object (including the ball) called the kinetic energy. The kinetic energy is defined to be K = ½mv2 where m is the mass of the object and v the speed. The SI units for kinetic energy are the same as for work, the joule (J). The kinetic energy of an object is always positive (or zero) it can never be negative. We can relate the kinetic energy of an object to the total work done on that object (or the work done by the net force) by W = K f - KI where Ki and Kf are the initial and final kinetic energies of the object. This is called the work-kinetic energy theorem and says that the total work done on an object is equal to the change in kinetic energy of that object. So when the boy does work on the ball he also increases the kinetic energy of the ball. Q. A pot plant falls from a ledge and drops 10 m to the ground below. How fast was it falling just before hitting the ground? A. We can solve this problem either by using our equations for free-fall or the work-kinetic energy theorem. Let’s us the later. The only force acting during free-fall is due to gravity F = mg Since this is constant and in the same direction as motion W = mg x The initial speed is 0 so Ki = 0. The work-kinetic energy theorem gives mg x = ½m( v 2f vi2 ) vf = = 2gx 2 (9.8 m.s-2 ) (10 m) = 14 m.s-2 36 Power Sometimes we are not interested in the work done by a force but the rate at which the work is done. For example the top speed of a car is determined by the rate at which the motor can do work. This quantity is called the power and the average power is defined to be P = W t where W is the work done in a time t. The instantaneous power is given by P = dW dt The SI units for power are the joule per second (J.s-1) but it is also used very often and so has a special name, the watt (W). 37 38