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Solving Problems with Energy Conservation, continued.

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Presentation on theme: "Solving Problems with Energy Conservation, continued."— Presentation transcript:

1 Solving Problems with Energy Conservation, continued

2 Conservation of Mechanical Energy   KE +  PE = 0 (Conservative forces ONLY!! ) or E = KE + PE = Constant For elastic (Spring) PE: PE elastic = (½)kx 2 KE 1 + PE 1 = KE 2 + PE 2  (½)m(v 1 ) 2 + (½)k(x 1 ) 2 = (½)m(v 2 ) 2 +(½)k(x 2 ) 2 x 1 = Initial compressed (or stretched) length x 2 = Final compressed (or stretched) length v 1 = Initial velocity, v 2 = Final velocity

3 Example 6-11: Toy Dart Gun E = (½)m(v 1 ) 2 + (½)k(x 1 ) 2 = (½)m(v 2 ) 2 + (½)k(x 2 ) 0 + ( ½ )(250)(0.06) 2 = (½)(0.1)(v 2 ) Gives: v 2 = 3 m/s Mechanical Energy Conservation A dart, mass m = 0.1 kg is pressed against the spring of a dart gun. The spring (constant k = 250 N/m) is compressed a distance x 1 = 0.06 m & released. The dart detaches from the spring it when reaches its natural length (x = 0). Calculate the speed v 2 it has at that point.

4 Example: Pole Vault Estimate the kinetic energy & the speed required for a 70-kg pole vaulter to just pass over a bar 5.0 m high. Assume the vaulter’s center of mass is initially 0.90 m off the ground & reaches its maximum height at the level of the bar itself.

5 Example 6-12: Two Kinds of PE Step 2: (b)  (c) (both gravity & spring PE)  (½)m(v 2 ) 2 + (½)k(y 2 ) 2 + mgy 2 = (½)m(v 3 ) 2 + (½)k(y 3 ) 2 + mgy 3 y 3 = Y = 0.15m, y 2 = 0  (½)m(v 2 ) 2 = (½)kY 2 - mgY Solve & get k = 1590 N/m ALTERNATE SOLUTION: (a)  (c) skipping (b) v 1 = 0 v 2 = ? v 3 = 0 m = 2.6 kg, h = 0.55 m Y = 0.15 m, k = ? A 2 step problem: Step 1: (a)  (b) (½)m(v 1 ) 2 + mgy 1 = (½)m(v 2 ) 2 + mgy 2 v 1 = 0, y 1 = h = 0.55 m, y 2 = 0 Gives: v 2 = 3.28 m/s

6 Example: Bungee Jump (a)  (c) directly!!  Δy + 15m  m = 75 kg, k = 50 N/m, y 2 = 0 v 1 = 0, v 2 = 0, y 1 = h = 15m +  y  y =? Mechanical Energy Conservation with both gravity & spring (elastic) PE  (½)m(v 1 ) 2 + mgy 1 = (½)m(v 2 ) 2 + mgy 2 + (½)k(  y) mg(15+  y) = (½)k(  y) 2 Quadratic Equation for  y: Solve & get  y = 40 m & -11 m (throw away negative value)

7 Other forms of energy; Energy Conservation In any process Total energy is neither created nor destroyed. Energy can be transformed from one form to another & from one body to another, but the total amount is constant.  Law of Conservation of Energy Again: Not exactly the same as the Principle of Conservation of Mechanical Energy, which holds for conservative forces only! This is a general Law!! Forms of energy besides mechanical: –Heat (conversion of heat to mech. energy & visa-versa) –Chemical, electrical, nuclear,..

8 The total energy is neither decreased nor increased in any process. Energy can be transformed from one form to another & from one body to another, but the total amount remains constant  Law of Conservation of Energy Again: Not exactly the same as the Principle of Conservation of Mechanical Energy, which holds for conservative forces only! This is a general Law!!

9 Sect. 6-9: Problems with Friction We had, in general: W NC =  KE +  PE W NC = Work done by all non-conservative forces  KE = Change in KE  PE = Change in PE (conservative forces only) Friction is a non-conservative force! So, if friction is present, we have (W NC  W fr ) W fr = Work done by friction Moving through a distance d, friction force F fr does work W fr = - F fr d

10 When friction is present, we have: W fr = -F fr d =  KE +  PE = KE 2 – KE 1 + PE 2 – PE 1 –Also now, KE + PE  Constant! –Instead, KE 1 + PE 1 + W fr = KE 2 + PE 2 or: KE 1 + PE 1 - F fr d = KE 2 + PE 2 For gravitational PE: ( ½) m(v 1 ) 2 + mgy 1 = ( ½) m(v 2 ) 2 + mgy 2 + F fr d For elastic or spring PE: (½)m(v 1 ) 2 + (½)k(x 1 ) 2 = (½)m(v 2 ) 2 + (½)k(x 2 ) 2 + F fr d

11 Example 6-13: Roller Coaster with Friction Calculate the work done by friction (the thermal energy produced) & calculate the average friction force on the car. m = 1000 kg, d = 400 m, y 1 = 40 m, y 2 = 25 m, v 1 = y 2 = 0, F fr = ? (½)m(v 1 ) 2 + mgy 1 = (½)m(v 2 ) 2 + mgy 2 + F fr d  F fr = 370 N A roller-coaster car, mass m = 1000 kg, reaches a vertical height of only y = 25 m on the second hill before coming to a momentary stop. It travels a total distance d = 400 m.

12 Sect. 6-10: Power Power  Rate at which work is done or rate at which energy is transformed: Average Power: P = (Work)/(Time) = (Energy)/(time) Instantaneous power: SI units: Joule/Second = Watt (W) 1 W = 1J/s British units: Horsepower (hp). 1hp = 746 W A side note: “Kilowatt-Hours” (from your power bill). Energy! 1 KWh = (10 3 Watt)  (3600 s) = 3.6  10 6 W s = 3.6  10 6 J

13 Example 6-14: Stair Climbing Power A 60-kg jogger runs up a long flight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m. a. Estimate the jogger’s power ss output in watts and horsepower. b. How much energy did this ss require?

14 Average Power Its often convenient to write power in terms of force & speed. For example, for a force F & displacement d in the same direction, we know that the work done is: W = F d So  F (d/t) = F v = average power v  Average speed of the object

15 Example 6-15: Power needs of a car Calculate the power required for a 1400-kg car to do the following: a. Climb a 10° hill (steep!) at a steady 80 km/h b. Accelerate on a level road from 90 to 110 km/h in 6.0 s Assume that the average retarding force on the car is F R = 700 N. a. ∑F x = 0 F – F R – mgsinθ = 0 F = F R + mgsinθ P = Fv l b. Now, θ = 0 ∑F x = ma F – F R = 0 v = v 0 + at P = Fv


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