Presentation on theme: "Aim: How can we solve problems dealing with the Law of Conservation of Energy? HW #10 due tomorrow Do Now: A 10 kg object free falls off the top of a 100."— Presentation transcript:
Aim: How can we solve problems dealing with the Law of Conservation of Energy? HW #10 due tomorrow Do Now: A 10 kg object free falls off the top of a 100 m tall building. At the top, calculate its potential and kinetic energies. U = mghK = ½mv 2 U = (10 kg)(9.8 m/s 2 )(100 m)K = ½(10 kg)(0 m/s) 2 U = 9,800 JK = 0 J HW #9 Answer Key
Conservation of Energy The total energy (E T ) of an object or system is constant Energy cannot be created or destroyed Energy can be changed from one form to another.
A Dropped Sphere What energy does it have when held off the ground? Potential Energy What happens to this PE as the ball drops? It becomes smaller and smaller Is the energy just disappearing? No!! It is being converted into KE (the object is speeding up)
100 m m = 10 kg At the top: PE = 9,800 J (PE = mgh) KE = 0 J (at rest) E T = 9,800 J (PE + KE = E T ) At the bottom: PE = 0 J (no height) E T = 9,800 J (total energy is constant!) KE = 9,800 J (PE + KE = E T ) If PE = 4,900 J, what is KE? KE = 4,900 J PE = 3,000 J, what is KE? KE = 6,800 J
1.The figure above shows a rough semicircular track whose ends are at a vertical height h. A block placed at point P at one end of the track is released from rest and slides past the bottom of the track. Which of the following is true of the height to which the block rises on the other side of the track? (A) It is equal to h/2 (B) It is equal to h/4 (C) It is equal to h/2 (D) It is equal to h (E) It is between zero and h; the exact height depends on how much energy is lost to friction. If no energy were lost, the answer would be (d) No calculator **1 minute**
2. If the potential energy is zero at point II, where will the kinetic and potential energies of the ball be equal? (A) At point II (B) At some point between II and III (C) At point III (D) At some point between III and IV (E) At point IV 3. The speed of the ball at point II is most nearly (A) 3.0 m/s (B) 4.5 m/s (C) 9.8 m/s (D) 14 m/s (E) 20 m/s A ball swings freely back and forth in an arc from point I to point IV, as shown at the right. Point II is the lowest point in the path, III is located 0.5 meter above II, and IV is 1 meter above II. Air resistance is negligible. ΔK = ΔU ½ mv 2 = mgh ½ v 2 = gh No calculator **2 minutes**
4. What is the kinetic energy of the rock just before it hits the ground? (A)mgh (B)½ mv 0 2 (C)½ mv 0 2 – mgh (D)½ mv mgh (E)mgh - ½ mv 0 2 A rock of mass m is thrown horizontally off a building from a height h, as shown above. The speed of the rock as it leaves the thrower’s hand at the edge of the building is v 0. No calculator **1 minute 15 sec**
a. Using the principle of conservation of energy, develop an expression for the speed of the child at the lowest point in the swing in terms of g, R, and cos o 5. A child of mass M holds onto a rope and steps off a platform. Assume that the initial speed of the child is zero. The rope has length R and negligible mass. The initial angle of the rope with the vertical is o, as shown in the drawing above. Calculator **8 minutes**
b. The tension in the rope at the lowest point is 1.5 times the weight of the child. Determine the value of cos o.
a. What is the tension T in the string? 6. Two 10 ‑ kilogram boxes are connected by a massless string that passes over a massless frictionless pulley as shown. The boxes remain at rest, with the one on the right hanging vertically and the one on the left 2.0 meters from the bottom of an inclined plane that makes an angle of 60° with the horizontal. The coefficients of kinetic friction and static friction between the Ieft ‑ hand box and the plane are 0.15 and 0.30, respectively. You may use g = 10 m/s 2, sin 60° = 0.87, and cos 60° = ΣF = 0 T – mg = 0 T = mg T = (10 kg)(10 m/s 2 ) T = 100 N Calculator **15 minutes**
b. On the diagram below, draw and label all the forces acting on the box that is on the plane. T N FfFf mg
c. Determine the magnitude of the frictional force acting on the box on the plane. ΣF = 0 T – F f – mgsinθ = 0 F f = T – mgsinθ F f = 100 N – (10 kg)(10 m/s 2 )(0.87) F f = 13 N
The string is then cut and the left ‑ hand box slides down the inclined plane. d.Determine the amount of mechanical energy that is converted into thermal energy during the slide to the bottom. W f = F f x W f = μ k Nx W f = μ k mgcosθx W f = (0.15)(10 kg)(10 m/s 2 )(0.5)(2 m) W f = 15 J e.Determine the kinetic energy of the left ‑ hand box when it reaches the bottom of the plane. U = K + W f K = U – W f K = mgh – 15 J K = mgxsin60° - 15 J K = (10 kg)(10 m/s 2 )(2 m)(0.87) – 15 J K = 159 J
7. A 5.0 ‑ kilogram monkey hangs initially at rest from two vines, A and B. as shown above. Each of the vines has length 10 meters and negligible mass. a. On the figure below, draw and label all of the forces acting on the monkey. (Do not resolve the forces into components, but do indicate their directions.) TATA TBTB mg Calculator **10 minutes**
b. Determine the tension in vine B while the monkey is at rest.
The monkey releases vine A and swings on vine B. Neglect air resistance. c. Determine the speed of the monkey as it passes through the lowest point of its first swing.
d. Determine the tension in vine B as the monkey passes through the lowest point of its first swing.