95.141 Mar 20, 2013 PHYSICS I Lecture 14 Exam II Wed Mar 27 Review Session Mon Mar 25 6-8:30 pm Ball 210 Course website: faculty.uml.edu/pchowdhury/95.141/ www.masteringphysics.com Course: UML95141SPRING2013 Lecture Capture h"p://echo360.uml.edu/chowdhury2013/physics1spring.html
Exam II Info Exam II Wed Mar 27 9:00-9:50am, OH 150. Exam II covers Chapters 4-8 Same format as Exam I Prior Examples of Exam II posted Ch. 4: Forces and Newton s Laws of motion Ch. 5: Using Newton s Laws, Friction, Uniform Circular motion Ch. 6: Universal Law of Gravitation, Kepler s Laws Ch. 7: Work & Energy Ch. 8: Conservation of Energy Exam Review Session Mon Mar 25 6:00 8:30 pm, Ball 210
Last Lecture Today Chapter 8 Conservative vs Nonconservative forces Potential Energy Conservation of Mechanical Energy Problems & Examples Chapter 8 Conservation of Energy with Dissipative forces Thermal Energy Gravitational Potential Energy Escape Velocity Power Problems & Examples
Conservation of Energy For a system where only conservative forces do work, energy is transformed between kinetic and potential Work Energy Principle W net =!K Relation between potential energy and work:!w net = "U!K = "!U!K +!U = 0 (K 2! K 1 )+ (U 2!U 1 ) = 0 K 2 +U 2 = K 1 +U 1 Define Mechanical Energy E 2 = E 1 = constant E = K +U Conservation of Mechanical Energy
Clicker Quiz Paul and Kathleen start from rest at the same time from the top of a frictionless water slides with different shapes. Who makes it to the bottom first? 1) Paul 2) Kathleen 3) both the same
Example Object starting from rest, sliding in valley h 1 2 3 h Ideal world ΔK + ΔU = 0 h 1 2 3 h Real world ΔK + ΔU 0 ΔK + ΔU + ΔE thermal = 0
Where does the Energy go? If friction acts on the object, then conservation of mechanical energy no longer holds. But does the energy just disappear? It is converted into heat! Thermal Energy The total energy is neither increased nor decreased in any process. Energy can be transformed from one form to another, and transferred from one object to another. New and improved law of energy conservation.
Example A 1-kg block slides down the plane shown below (θ=30º). The block starts with v 1 =6.0 m/s. A) If there is no friction, what the speed of the block be at point 2? 1!K +!U gravity = 0 h=10m! 2 ( 1 mv 2 2 2! 1 mv 2 2 1 )+ (0! mgh) = 0 v 2 2 = v 1 2 + 2gh v 1 =15 m s
Example A 1-kg block slides down the plane shown below (θ=30º). The block starts with v 1 =6.0 m/s. B) The block reaches point 2 with v 2 =10 m/s. How much thermal energy is generated? 1!K +!U gravity +!E thermal = 0 h=10m 2!K = 32 J!U = "98 J!E thermal = +66 J
Example A 1-kg block slides down the plane shown below (θ=30º). The block starts with v 1 =6.0 m/s. C) The block reaches point 2 with v 2 =10 m/s. What is µ k?!e thermal = "W friction 1 W friction =!µ k (mgcos!)d h=10m 2 h W friction =!µ k (mgcos!)( sin! ) W friction =! µ kmgh tan! =!66J µ k = 0.39
Example 1 Can also approach this problem from Work-Energy perspective Net work done on mass = change in kinetic energy h=10m 2
Gravitational Potential Energy We have so far written the change in gravitational potential energy as ΔU = mgδy. Near the surface of the Earth, this is a good approximation. However, remember, the exact definition for the change in gravitational potential energy is:!u =U 2 "U 1 = "W g = "# F! g d! " We know that F g changes as a function of distance! 2 1
Exact Gravitational Potential Energy! F g =!G mm E r 2 ˆr W g = 2!! F g d! " 1 Can define U g =!G mm E r Check that F g =! du g dr!u = " GmM E r 2 + GmM E r 1 =U 2 "U 1 U=0 is now chosen to be at infinity instead of at R E Conservation of Energy for gravity (exact) 1 2 mv 2 1! GmM E = 1 r 1 2 mv 2 2! GmM E r 2 = constant
Escape Velocity We are used to thinking that an object shot up from the Earth s surface will return to Earth. Is this always the case? Say we shoot a rocket with mass m, v o = 2!10 4 m/s = 20 km/s R E = 6.38!10 6 m M E = 5.98!10 24 kg G = 6.67!10 "11 Nm 2 / kg 2 E 0 = KE + PE = 1 2 mv 2 0! GmM E R E E 0 = m!1.37!10 8 J E 0 = KE + PE = 1 2 mv 2 f! GmM E R " v f =1.66!10 4 m/s = 16.6 km/s
Escape Velocity What is the lowest speed we can launch a rocket and have it never come back? E 0 = KE + PE = 1 2 mv 2 esc! GmM E R E v esc = 2GM E R E = 2gR E v esc =11.2 km / s This is called escape velocity
Power Power The rate at which work is done The work done divided by the time it takes to do the work. The rate at which energy is transformed. Work done by a force can be transformed into kinetic energy, potential energy, thermal energy, units Watts = Joules/sec P = dw dt = de dt P = dw dt = d(! F d! l ) dt =! F d! l dt =! F! v
Example How much power does it take a 50-kg runner to run up a 5 m high hill in 20 s? P =122.5 W What about in 10 s? P = 245 W
Apollo 13 CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 14
Clicker Quiz Approximately how much thermal energy is generated upon reentry? A) ~kj s B) ~MJ s C) 100 s of MJ s D) ~GJ s E) 100 s of GJ s
Apollo 13 How much Thermal Energy is generated in reentry? What is the average power generated in reentry? v Altitude M v o f Altitude Δt = 10,743.6 module o = = 134 f = 121,920m 4300kg m s = 300s m s = 7,620m M G R E Earth = 5.98 10 = 6.67 10 11 24 = 6.38 10 kg Nm 6 2 m kg 2
Solve for Energies Initial Mech. Energy U( r) GM = r E = 1.57 10 m 10 J Final Mech. Energy = 2.68 10 11 11 E th = 2.53 10 J J Power 8 P = 8.43 10 W = 843 MW
Hoover Dam Maximum Capacity: ~2 GW Height: 221.4m What is the water flow required to generate the max capacity for Hoover Dam, if turbines are 60% efficient and no dissipative forces? Solve for flow of 1 kg/s ΔU = ΔK = Ewater = 2165. 8 J E turbine = 1300 J 6 kg Flow =1.66 10 s
Verasca Dam in Switzerland Height: 220 m CHOWDHURY 95.141 PHYSICS I SPRING 2013 LECTURE 14
Summary Conservation of Energy with Dissipative forces Thermal Energy Gravitational Potential Energy Problems & Examples