Solving Problems with Energy Conservation Conservation of Mechanical
Solving Problems with Energy Conservation
Conservation of Mechanical Energy KE + PE = 0 (Conservative forces ONLY!! ) or E = KE + PE = Constant • For elastic (Spring) PE: PEelastic = (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
Example: Toy Dart Gun Mechanical Energy Conservation E= 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. (½)m(v 1)2 + (½)k(x 1)2 = (½)m(v 2)2 + (½)k(x 2)2 0 + ( ½ )(250)(0. 06)2 = (½)(0. 1)(v 2)2 + 0 Gives: v 2 = 3 m/s
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.
Example: Two Kinds of PE v 1 = 0 m = 2. 6 kg, h = 0. 55 m Y = 0. 15 m, k = ? v 2 = ? 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 v 3 = 0 Gives: v 2 = 3. 28 m/s 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 = 0. 15 m, y 2 = 0 (½)m(v 2)2 = (½)k. Y 2 – mg Solve & get k = 1590 N/m ALTERNATE SOLUTION: (a) (c) skipping (b)
Example: Bungee Jump m = 75 kg, k = 50 N/m, y 2 = 0 v 1 = 0, v 2 = 0, y 1 = h = 15 m + 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)2 = 0 + mg(15+ y) = 0 + (½)k( y)2 Quadratic Equation for y: Solve & get y = 40 m & -11 m (throw away negative value) Δy = + 15 m (a) (c) directly!!
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, . .
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!!
Problems with Friction • We had, in general: WNC = KE + PE WNC = 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 (WNC Wfr) Wfr = Work done by friction Moving through a distance d, friction force Ffr does work Wfr = - Ffrd
When friction is present, we have: Wfr = -Ffrd = KE + PE = KE 2 – KE 1 + PE 2 – PE 1 Also now, KE + PE Constant! Instead, or: KE 1 + PE 1+ Wfr = KE 2+ PE 2 KE 1 + PE 1 - Ffrd = KE 2+ PE 2 • For gravitational PE: (½)m(v 1)2 + mgy 1 = (½)m(v 2)2 + mgy 2 + Ffrd • 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 + Ffrd
Example: Roller Coaster with Friction 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. 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, Ffr = ? (½)m(v 1)2 + mgy 1 = (½)m(v 2)2 + mgy 2 + Ffrd Ffr= 370 N
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 = 1 J/s British units: Horsepower (hp). 1 hp = 746 W A side note: “Kilowatt-Hours” (from your power bill). Energy! 1 KWh = (103 Watt) (3600 s) = 3. 6 106 W s = 3. 6 106 J
Example: 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?
• 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=Fd So F (d/t) = F v = average power v Average speed of the object
Example: 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 FR = 700 N. a. ∑Fx = 0 F – FR – mgsinθ = 0 F = FR + mgsinθ P = Fv l b. Now, θ = 0 ∑Fx = ma F – F R= 0 v = v 0 + at P = Fv
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