KE = ½mv 2 PE = mgh W = Fdcosθ THINK ENERGY! (KE F + PE F ) = (KE 0 + PE 0 ) + W NC. Tues Oct 6 Assign 7 Fri Pre-class Thursday

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1 Tues Oct 6 Assign 7 Fri Pre-class Thursday Conservation of Energy Work, KE, PE, Mech Energy Power To conserve total energy means that the total energy is constant or stays the same. With Work, we now have way to talk about getting started and coming to a stop. W = F d cosθ (KE F + PE F ) = (KE 0 + PE 0 ) + W NC m,v m,h F,d THINK ENERGY! KE = ½mv 2 PE = mgh W = Fdcosθ PhET

2 Solving Problems Energies? v? h? No time? (clues) Draw - visualize Decide on initial and final events Choose reference level for PE Any W NC? W NC = ΔE or (KE i + PE i ) + W NC = (KE f + PE f )

3 Suppose I drop a 0.2kg mass ball from a height 3.0m above the floor. What is the speed of the ball just before it hits the floor? Initial: Release ball Final: Just before hit floor Reference: floor h=0 h=3m E F = E I + W NC No air resistance, etc so Total Mech E stays the same. (W NC =0) KE F + PE F = KE I + PE I ½mv = 0 + mgh Cancel m gh = ½v 2 2(9.8 m/s 2 )(3m) =v 2 h=0m v=7.7 m/s

4 3 balls of the same mass are thrown from a cliff, all with a speed of 25m/s. A is thrown upward at an angle of 45. B is thrown horizontally. C is thrown downward at an angle of 45. Which one is traveling fastest just before it hits the ground? 1. A 2. B 3. C 4. All three are traveling the same speed Same initial KE, same change in PE, same increase in KE. Does the mass matter? Not in this case. It would cancel. E F = E 0 + W NC KE F + PE F = KE I + PE I ½mv = ½m(25m/s) 2 + mgh

5 Bowling Ball Pendulum Forces? W TENSION =? Energy Lost? Gained?

6 A block starts from rest and slides down a frictionless ramp. How does the speed at Point B compare to the speed at point G? A B C D E F G H (1) v B < v G (2) v B = v G (3) v B > v G PE same at points B, E and G. If W NC = 0, then E TOT same and therefore KE same. (KE F + PE F ) = (KE 0 + PE 0 ) + W NC Same

7 A particle, starting from point A in the drawing, is projected down the curved runway. Upon leaving the runway at point B, the particle is traveling straight upward and reaches a height of 4.00m above the floor before falling back down. Ignore friction and air resistance. Find the speed of the particle at point A. Initial: Leave point A Final: Max Height Ref Level: Table Top (KE F + PE F ) = (KE 0 + PE 0 ) + W NC No non-conservative forces (0 + mg(4.0 m)) = (½mv mg(3.0 m)) + 0 Can cancel mass, rearrange and solve (9.8 m/s 2 )(4.0 m 3.0 m) = ½v 0 2 v 0 = 4.4m/s

8 Four identical balls roll off four tracks. The tracks are of the same height but different shapes, as shown. For which track is the ball moving the fastest when it leaves the track? (1) Yellow (2) Red (3) Green (4) Blue (5) All the same Start at same height, end at same height, start with same KE Same initial total energy No loss of mechanical energy Same loss in Potential energy So same final kinetic energy If all leave track at same angle, same distance before hit the ground

9 Four identical balls roll off four tracks. The tracks are of the same height but different shapes, as shown. For which track does the trip from start to finish take the least time? (1) Yellow (2) Red (3) Green (4) Blue (5) All the same Greatest acceleration early on Note: Gain ability to handle strange shapes and paths LOSE time information and direction information. Get the magnitude of velocity, but not direction.

10 An 80-kg stuntman starts at rest and slides down a roof, flies through the air, and lands on a large pad, which compresses 1.2m in order to bring the stuntman to a stop. Assume it is an icy day and the roof is frictionless. Ignore air resistance. What is the average force on the stuntman due to the pad? KE PE E TOT Top 0 mg(4m) 4mg Top of Pad ½mv 2 0 ½mv 2 Rest 0 mg(-1.2m) -1.2mg Initial: Top Final: Rest Ref Level: Top of uncompressed Pad (KE F + PE F ) = (KE 0 + PE 0 ) + W NC (0+( 1.2)mg) = (0 + 4mg) + F*d*cos(180º) 1.2mg = 4mg + F*(1.2)*(-1) 1.2F = 5.2mg F = 3400 N

11 An 80-kg stuntman slides down a roof, flies through the air, and lands on a large pad, which compresses 1.2m in order to bring the stuntman to a stop. Assume it is an icy day and the roof is frictionless. Ignore air resistance. What is the speed of the stuntman just (1) 4.4 m/s (2) 4.8 m/s before he hits the pad? (3) 6.3 m/s (4) 7.7 m/s (5) 8.8 m/s (6) 78 m/s Initial: top, Final: just before hit No energy lost (KE F + PE F ) = (KE 0 + PE 0 ) + W NC (½mv 2 + 0) = (0 + 4mg) + 0 v = 8.85 m/s

12 At the same time that Stuntman A slides down the roof (starting from rest), Stuntman B steps off a roof (starting at the same height) on the other side of the street. Assume a 'no-friction' roof and negligible air resistance. Which Stuntman is traveling faster when they make contact with the pad? (1) A (2) B (3) both have the same speed Start at same height and initial speed, end at same height, same energy lost due to pad, no friction

13 At the same time that Stuntman A slides down the roof, Stuntman B steps off a roof (starting at the same height) on the other side of the street. Which Stuntman makes contact with the pad first? (1) A (2) B (3) both make contact at the same time Conservation of energy doesn t help us here. A is not in free fall until leaves roof, so for a while acceleration smaller than B s acceleration.

14 Suppose the roof on which A is sliding is NOT frictionless. Which Stuntman is traveling faster when they make contact with the pad? (1) A (2) B (3) both have the same speed Work due to friction reduces total energy of A.

15 A crate is given a kick by a spring such that it has 50J of kinetic energy. It travels along a straight frictionless track until it hits a 10m stretch of sandpaper. For this stretch, the frictional force is 4N. The track becomes frictionless again after the sandpaper. At which point or points does the box has a velocity of zero? 1. A (1/4) 2. B (1/2) A B C D E 3. C (3/4) 4. D (Flat after friction) 5. E (Highest point on slope after friction) 6. E and C 7. E and B 8. E and A Can lose KE due to Work done by frictional force or by transforming to PE. Total work done by friction = F*d*cos(180) = 4*10*(-1) = -40J Has 10J left. Climbs up slope, stops momentarily, then comes back. If W=F*d*cos(180); -10 = 4*d*(-1); d= 2.5m

16 Conservation of Energy in General So far, just mechanical energy Other Energy (OE) types: Thermal, Electrical, Chemical, Nuclear, Sound, Light,... "Energy can neither be created nor destroyed, but only converted from one form to another. KE i + PE i + W NC + OE i = KE f + PE f + OE f

17 Power Rate at which work is done Rate at which energy is expended J/s watt (W) P = Work Time P = Change in Energy Time

18 A motor has a power rating of 500W. How much energy can the motor provide when run for 20 s? (1) 20 J (2) 25 J (3) 500 J (4) 10,000 J 500 W is 500 J/s. Every second you get 500 J 500 J/s * 20s = 10,000 J

19 Work, Energy, and Power in Humans Efficiency (eff) = Useful Energy or Work Output / total Energy Input

20 What is the minimum energy needed for you to climb a 20m-tall cliff? Assume a mass of 70kg. Also assume that you are at rest before and after the climb. (1) 14.3 J (2) 140 J (3) 196 J (4) 463 J (5) 686 J (6) J KE i + PE i + W NC = KE f + PE f Pick reference at base 0J + 0J + W NC = 0 J + mgh W NC = (70kg) (9.8m/s 2 )(20m) = 13700J

21 What is the minimum power output if this climb is accomplished over a time of 3 minutes? (1) 76.2 W (2) 2380 W (3) 4570 W (4) 8290 W (5) W (6) W Power = Energy or Work / time = J / 180 s = 76.2 W

22 If your body is only 0.20 (20%) efficient at creating useful energy from your energy input (food), what would be the energy input needed for the climb? (1) 2740 J (2) J (3) J (4) J (5) J Eff = Useful Energy / Energy Input Energy Input = Useful Energy / eff = J / 0.20 = J How much non-useful energy would be produced? (typically in the form of thermal energy) Energy input = Energy useful + Energy not useful E NOT USEFUL = J J = J

23 Stuntman Look for similarities! Diver Fool Stop Stop

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