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Examples: Mechanical Energy Conservation

Examples: Mechanical Energy Conservation. Example 8.3 – Spring Loaded Cork Gun.

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Examples: Mechanical Energy Conservation

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  1. Examples: Mechanical Energy Conservation

  2. Example 8.3 – Spring Loaded Cork Gun • Ball, mass m = 35 g = 0.035 kg in popgun is shot straight up with spring of unknown constant k. Spring is compressed yA = - 0.12 m, below relaxed level, yB = 0. Ball gets to a max height yC = 20.0 m above relaxed end of spring. (A) If no friction, find spring constant k. (B) Find speed of ball at point B. • Ball starts from rest. Speeds up as spring pushes against it. As it leaves gun, gravity slows it down. System = ball, gun, Earth. • Conservative forces are acting, so use Conservation of Mechanical Energy Initial kinetic energy K = 0. Choose gravitational potential energy Ug = 0 where ball leaves gun. Also elastic potential energy Ue= 0 there. At max height, again have K = 0. Choose Note: Need two types of potential energy!

  3. For entire trip of ball, Mechanical Energy is Conserved!! or: KA + UA = KB + UB = KC + UC. At each point, U = Ug + Ue so, KA+ UgA+ UeA = KB+ UgB + UeB = KC + UgC+UeC (A) To find spring constant k, use: KA+ UgA+ UeA = KC + UgC+UeC or, 0 + mgyA + (½)k(yA)2 = 0 + mgyC + 0, giving k = [2mg(yC – yA)/(yA)2] = 958 N/m (B) To find ball’s speed at point B, use: KA+ UgA+ UeA = KB + UgB+UeB or, (½)m(vB)2 + 0 + 0 = 0 + mgyA + (½)k(yA)2 , giving (vB)2 = [k(yA)2/m] + 2gyA; or, (vB) =19.8 m/s

  4. Example: Toy Dart Gun • Mechanical energy conservation!  (½)m(v1)2 + (½)k(x1)2 = (½)m(v2)2 + (½)k(x2)2 • Speed when ? leaves gun? x1 = 0.06 m, v1 = 0, m = 0.1 kg, k = 250 N/m x2 = 0, v2 = ? Find: v2 = 3 m/s ?

  5. v1 = 0 Example: Two Kinds of PE v2 = ? v3 = 0 m =2.6 kg h =0.55 m Y = 0.15 m k = ? A two step problem:STEP 1: (a)  (b)  (½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 v1 = 0, y1 = h = 0.55 m, y2 =0 . Find: v2 = 3.28 m/s STEP 2: (b)  (c) (both gravity & spring PE)  (½)m(v2)2+(½)k(y2)2+mgy2 = (½)m(v3)2 + (½)k(y3)2 + mgy3 y3 = Y = 0.15m, y2 = 0  (½)m(v2)2 = (½)kY2 - mgY Solve for k & get k = 1590 N/m ALTERNATE SOLUTION: (a)  (c) skipping (b)

  6. Section 8.3: Problems with Friction • We had, in general: WNC = K + U WNC = Work done by non-conservative forces K = Change in KE U = Change in PE (conservative forces) • Friction is a non-conservative force! So, if friction is present, we have (WNC  Wf) Wf = Work done by friction In moving through a distance d, force of kinetics friction fkdoes work Wf = - fkd

  7. When friction is present, we have: Wf= -fkd = K + U = Kf – Ki + Uf – Ui • Also now, K + U  Constant! • Instead, Ki + Ui+ Wf = Kf + Uf OR: Ki + Ui - fkd = Kf+ Uf • For gravitational PE: (½)m(vi)2 + mgyi = (½)m(vf)2 + mgyf + fkd • For elastic or spring PE: (½)m(vi)2 + (½)k(xi)2 = (½)m(vf)2 + (½)k(xf)2 + fkd

  8. Example: Roller Coaster with Friction m=1000 kg, d=400 m, y1=40 m, y2= 25 m, v1= y2 = 0, fk= ? (½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 + fkd  fk= 370 N

  9. Ex. 8.4 – Block Pulled on Rough Surface • A block, mass m = 6 kg, is pulled by constant horizontal force F = 12 N. over a rough horizontal surface. Kinetic friction coefficient μk = 0.15. Moves a distance Δx = 3 m. Find the final speed.

  10. Example 8.6 – Block – Spring System • A mass m = 1.6 kg, is attached to ideal spring of constant k = 1,000 N/m. Spring is compressed x = - 2.0 cm = - 2  10-2 m& is released from rest. (A)Find the speed at x = 0 if there is no friction. (B) Find the speed at x = 0 if there is a constant friction force fk = 4 N.

  11. Ex. 8.7 – Crate Sliding Down a Ramp • Crate, mass m = 3.0 kg, starts from rest at height yi = 0.5 m & slides down a ramp, length • d = 1.0 m & incline angle • θ= 30°. Constant friction force fk = 5 N. Continues to move on horizontal surface after. • (A)Find the speed at the bottom. • (B)Assuming the same friction force, find the distance on then horizontal surface that the crate moves after it leaves the ramp.

  12. Ex. 8.8 – Block-Spring Collision • Block, mass m = 0.8 kg, gets initial velocity vA = 1.2 m/s to right. Collides with spring with constant k = 50 N/m. • (A)If no friction, find the maximum compression distance xmax of spring after collision. • (B)There is a constant friction force fk between block & surface. Coefficient of friction is μk = 0.5. Find the maximum compression distance xC now.

  13. Ex. 8.9 – Connected Blocks in Motion • Two blocks, masses m1& m2, are connected by spring of constant k. m1moves on horizontal surface with friction. Released from rest when spring is relaxed when m2at height h above floor. Eventually stops when m2is on floor. Calculate the kinetic friction coefficient μkbetween m1& table.

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