Physics 23 Notes Chapter 6 Part Two

Transcription

1 Physics 23 Notes Chapter 6 Part Two Dr. Alward Conservation of Energy Object moves freely upward under the influence of Earth only. Its acceleration is a = -g. v 2 = vo 2 + 2ax = vo 2-2g (h-ho) = vo 2-2gh + 2gho Multiple both sides by ½ m: ½ mv 2 = ½ mvo 2 - mgh + mgho ½ mv 2 + mgh = ½ mvo 2 + mgho K = ½ mv 2 Ko = ½ mvo 2 U = mgh Uo = mgho (final kinetic energy) (initial kinetic energy) (final potential energy) (initial potential energy) K + U = Ko + Uo E = Eo E = final total energy Eo = initial total energy 1

2 Example A: A ball of mass m = 10 kg located 20 meters above the ground is dropped. Assume the reference elevation is the ground. (a) What will be its kinetic energy when its potential energy is 1200 J? (a) Eo = Ko + Uo = (9.8)20 = 1960 J K = 1960 K = 760 J (b) ½ (10)v 2 = 760 v = m/s (b) What will be its speed at that point? Example B: The basketball about to fall into the basket is 2.2 meters above the elevation from which it was launched. What is its speed? Let the launching location be the reference elevation. ½mv 2 + mgh = ½ mvo 2 + mgho ½ mv 2 +mg (2.2) = ½ m (7.0) 2 + mg (0) v = 2.42 m/s 2

3 Example A: A 5.0 kg object at Point A is initially moving at 4.0 m/s on a frictionless hill. What will be its speed when it reaches point C? Solve problem using two different choices of a reference elevation. First Choice: Point B is zero elevation E = Eo ½ mv 2 + mg (15) = ½ m (4.0) 2 + mg (20) v = m/s Second Choice: Point C is zero elevation ½ mv = ½ m (4.0) 2 + mg (5) v = m/s Non-Conservative Work Work done by air resistance or friction doesn t allow the total energy to be conserved, because some of the kinetic energy is transformed into heat energy which exits the system; we speak of that work as being non-conservative work, and symbolize that work as WNC. E = Eo + WNC Example B: Suppose in the previous example, where the final speed was m/s, the actual speed is 9.24 m/s, instead. What would have been the work done by air resistance? Assume the object s mass is 2.0 kg. WNC = E - Eo = [K + U] - [Ko + Uo] = [½ (2.0)(9.24) (9.8) (25)] - [½ (2.0)(4.0) (9.8)(30)] = J Notice that the non-conservative work done is a negative number, as expected, because the frictional force points in a direction that is opposite to the direction of motion. 3

4 Work Creates Potential Energy A person pushes upward with a force P = mg that matches Earth s pull downward. That pushing force, times the height h, equals the work done by the pusher. W = Fx cos θ F = mg x = h θ = 0 o W = (mg) h cos 0 o = mgh mgh = W U = W The gravitational potential energy of an object at a certain elevation equals the work an agent had to do to raise it there from the reference level. 4

5 Spring Constants Spring constant: Symbol: k Units: N/m The spring constant of a spring is the number of newtons of force that must be applied to the spring to change the length of the spring by one meter: Example A: A 200-N force is required to stretch a spring by 0.40 m. What is the spring s spring constant? k = F/x = 200 N/0.40 m = 500 N/m Example B: How much force is necessary to stretch a 900 N/m spring by 0.20 m? F = kx = 900 N/m (0.20 m) = 180 N What average force was applied? F = ½ ( ) = 90 N 5

6 Spring Potential Energy Example: k = 2000 N/m A spring has a spring constant of 2000 N/m. What is its potential energy when it s stretched by 0.40 m, and compressed by 0.40 m? The force stretching the spring varies, so the work it does in stretching (or compressing) the spring by the amount x is the average force, times distance: W = x = ½ (0 + kx) x = ½ kx 2 Recall from the study of gravitational potential energy that potential energy is the work done to configure the system, we note therefore that U = W = ½ kx 2 Stretched: U = ½ (2000) (0.40) 2 = 160 J Compressed: U = ½ (2000)(-0.40) 2 = 160 J 6

7 Spring-Mass Systems Example: An object of mass m = 20.0 kg is attached to a spring lying on a frictionless tabletop. The other end of the spring is attached to a clamp on the table. The system was set in motion by pulling the object to the right and releasing it. At some moment, its kinetic energy is 500 J, while its spring potential energy is 900 J. What will be the speed of the object later, when the spring s potential energy is 300 J? Eo = Ko + Uo = = 1400 J E = Eo K + U = 1400 J ½ (20) v = 1400 v = m/s 7

8 Example: One end of a spring whose spring constant is 200 N/m is attached to the ceiling. At the other end is attached a 2.0 kg object, held there without allowing the spring to be stretched. The object is then released and allowed to fall while stretching the spring. How far below the object s initial location will the object reach before starting to move back upward? Call that distance x. Note that the final elevation of the mass is a negative number, -x, because the object is below the reference level. (See the second term on the left in the third equation below.) E = Eo K + U grav + U spring = Ko + Uo grav + Uo spring ½ (2)(0) 2 + 2(9.8)(-x) + ½ (200)x 2 = ½ (2)(0) 2 + 2(9.8)(0) + ½ (200)(0) 2 = 0 x = 0.20 m One would get the same result choosing, say, the bottom of the stretched spring as the reference level. 8