How does the total energy of the cart change as it goes down the inclined plane?

Transcription

1 Experiment 6 Conservation of Energy and the Work-Energy Theorem In this experiment you will explore the principle of conservation of mechanical energy. You will see that gravitational energy can be converted into kinetic energy. Recall that the gravitational potential energy of an object with mass m situated at a height h from the floor (taken as the zero of energy) is given by U = mgh where g is the acceleration due to gravity. Also, the kinetic energy of an object with mass m and moving with a velocity v is given by K = 1/mv. Objectives: 1. Understand the principle of the conservation of energy.. Apply the principle to situations involving gravitational potential energy and kinetic energy. 3. Make theoretical predictions of the final velocity of a cart after it rolls down an inclined plane using the principle of the conservation of energy. 4. Verify the work-energy theorem. Compare your predictions with experimental results. Hypothesis: How does the gravitational potential energy of a cart at the top of an inclined plane compare to the kinetic energy of the cart at the bottom of the inclined plane? How does the total energy of the cart change as it goes down the inclined plane? How does the work done on an object relate to the change in its kinetic energy? Theory: The mechanical energy E of a system is the sum of its potential energy U and the kinetic energy K of the objects within it. In this experiment we will examine what happens to this mechanical energy when only conservative forces (like the gravitational force) cause energy transfers within the system i.e. when frictional and drag forces do not act on the objects in the system. Also, we shall assume that the system is isolated from its environment i.e. no external force from an object outside the system causes energy changes inside the system. When a conservative force does work W on an object within the system, one of these energies increases exactly as much as the other decreases. 1

2 In an isolated system where only conservative forces cause energy changes, the kinetic energy and potential energy can change, but their sum, the mechanical energy E of the system, cannot change. This result is called the principle of conservation of mechanical energy. We can write this principle in one more form as follows: E mech = K + U = 0 The principle of conservation of mechanical energy allows us to solve problems that would be quite difficult to solve using only Newton s laws. When the mechanical energy of a system is conserved, we can relate the sum of the kinetic energy and potential energy at any one instant to that at another instant without considering the intermediate motion and without finding the work done by forces involved. h v Figure 1 above shows a cart stationary at the top of a frictionless track at a height h from the bottom (where we assume the gravitational potential energy is zero). The mechanical energy of the cart at this point is purely gravitational i.e. E = U = mgh. At the bottom of the track the cart has acquired a velocity and its gravitational energy is zero. The mechanical energy of the cart now is purely kinetic. i.e. E = K = ½mv. At any instant on the incline, the energy of the cart would be part potential and part kinetic. The total energy of the cart however would be the same in all three positions. Work-Energy Theorem When a body with mass m moves under the influence of a constant external force F parallel to its displacement d we say that the force has done some work. The amount of work done is given by: W = Fd

3 If the force applied is not along the direction of displacement of the object we need to take that component of the force along the direction of displacement to calculate the work done the applied force. In the Figure above, this force would be Fcos(α). Hence the work done in moving the block a distance d above is given by: W = Fd cos(α) The general expression for the work done by a constant force F that causes a displacement s is given by: F W = F. s α If the applied force is not constant during the displacement then the work done by the force needs to be calculated as follows: W = x f x i Fdx where F is the applied force along the x axis and x i and x f are the initial and final positions of the mass. The work-energy theorem states that if a mass m, moves along the x axis under the influence of an external force F that points along that axis, then the work done by this force on the mass m as it moves from an initial point x i to a final point x f along the x axis is equal to the change of the kinetic energy of the mass between the two points. This can be written mathematically as: W = K f K i = K If on a horizontal frictionless track a hanging mass pulls a cart that moves along the track, then the work done on the cart must equal the change in the kinetic energy of the cart. We will verify this theorem in the Laboratory. d 3

4 Exercises 1. You drop a.0 kg. text book from a height of 10 m from the ground to a friend who stands on the ground as shown in the Figure. (a) If the potential energy is taken as being zero at ground level, then what is the potential energy of the book when you release it? (b) What is the kinetic energy of the book just before your friend catches it in her outstretched hands (Hand level), which are 1. m above ground level. (c) How fast is the book moving as it is caught? Book 10 m Hand level 1. m Solution: (a) U = mgh = * 9.8 * 10 = 196 J (b) Potential energy of the book at hand level is mgh = *9.8*1. = 9.4 J Since the total energy (196 J) is conserved then the kinetic energy of the book is 196 J 9.4 J = J (c) K = J = ½mv.or v = 1.9 m/s 4

5 . A frictionless roller coaster car tops the first hill in the Figure with speed v o. What is its speed at (a) point A, (b) point B and (c) point C? (d) How high will it go on the last hill, which is too high for it to cross? v o A B h h h/ C Solution: Remember the total mechanical energy of the car is E = U + K = mgh + ½mv o and that this energy is conserved. (a) v o (b) E = mg(h/) + ½mv. Or mgh + ½mv o = mg(h/) + ½mv v o Solving for v we get v = gh + (c) E = ½mv. Or mgh + ½mv o = ½mv Solving for v we get v = gh + v o (d) At the highest point the roller coaster will come to a halt momentarily. At this height (H) the potential energy will be the same as the kinetic energy it had at point C. i.e. Use the velocity calculated in part c to find the kinetic energy at point C. 1 m( gh + vo ) = mgh vo H = h + g

6 3. A.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown below. How much work is done by the force as the block moves from the origin to 8.0 m? 1 10 Force (N) 0 - Solution: Position (m) Since the direction of the applied force is along the displacement of the block the work done (W = Fd) would just be the area under the curve shown above. We can calculate this area in four parts corresponding to the four section A,B,C and D as seen in the Figure above. Part A: W = 10 * = 0 J Part B: Here we need the slope of this section and the form of the equation. viz. we need to know how the force varies with x (i.e. F(x) ) in this region i.e. < x < 4. F(x)= -x + 0 < x < 4 Hence the work done in this section is 4 W = ( x + 0) dx = 10J Part C: No work is done here as F = 0. Part D: Here we need the slope of this section and the form of the equation. viz. we need to know how the force varies with x (i.e. F(x) ) in this region i.e. 6 < x < 8. F ( x) = x < x < 8 Hence the work done in this section is: 8 W = ( x + 6) dx = J 6 Hence the total work done is 0 J + 10 J - J = J 6