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Examples and Hints in Chapter 7

Examples and Hints in Chapter 7. 8 m . 10 m. Wild Monkey III.

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Examples and Hints in Chapter 7

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  1. Examples and Hints in Chapter 7

  2. 8 m 10 m Wild Monkey III Tarzan (m =100 kg) grabs a vine to swing to cross a chasm. He starts a from a cliff face that is 10 m taller than the opposite side and his vine is 8 m long. The vine will break if its tension exceeds 1200 N. Can Tarzan make the swing without the vine breaking?

  3. Draw a Free Body diagram Centripetal force, Mv2/R Tension, T At bottom of rope: Weight of Tarzan=100*9.8 =980 N

  4. Use the Work Energy Thm: DU+DK-Wother=0 • There are no non-conservative forces so DU+DK=0 Or K1+U1=K2+U2 Initially, Tarzan at rest (K1=0) but 10 m above final position (U1=mgy=980*10=9800J) Finally, Tarzan at lower potential (U2=0) but in full “swing” (K2= ½ mv2=0.5*100*v2) So 9800 = 0.5*100*v2 196=v2

  5. So

  6. One final question might be at what height does the rope snap?

  7. Bad Coaster A 1000 kg rollercoaster starts with a velocity of 10 m/s from the top of a track that is 65 m high. If frictional forces do -400 kJ of work on it as it approaches the top of the loop-de-loop which has a radius of 10 m, is the coaster able to make the loop? V1= 10 m/s R=10 m 65 m

  8. Draw a free body diagram at the top of the loop n, normal force (because upside down) Weight of coaster, mg=1000*9.8=9800

  9. Use the Work Energy Thm: DU+DK-Wother=0 • U1=mg*65=9800*65=637 kJ • K1= ½ *1000*10^2=50 kJ • U2= mg*2*R=9800*20=196 kJ • K2= ½ *1000*v2 • U1+K1-Wfriction=K2+U2 • 637 kJ+50 kJ-400 kJ=500v2 + 196 kJ • 91=500v2 • V2=0.182

  10. So

  11. 2 kg 4 m 53.10 Problem 7.74 A 2.00 kg package is released on a 53.10 incline, 4.00 m from a long spring with a force constant of 120 N/m that is attached to the bottom of an incline. The coefficients of friction between the package and the incline is ms=0.4 and mk=0.2. The mass of the spring is negligible. • What is the speed of the package just before it reaches the spring? • What is the maximum compression of the spring?

  12. 2 kg 4 m 53.10 Wgravity=2*9.8*4*cos(37.90) Wgravity=62.72J Mg cos(530) 5 37.90 4 f=mk*mg*cos(530) f=0.2*2*9.8*.6 f=2.354 N Wfriction=2.354*4=9.408 J 3 mg

  13. Use Work Energy Net Work=Wgravity-Wfriction =62.72-9.408=53.3 J DK=Net Work ½ mv2 – 0=53.3 V=7.3 m/s

  14. Part b) Use Work Energy -DU=Wnet Wfriction=2.354*(4+d)=9.408+2.354*d Wgravity=15.67*(4+d)=62.69+15.67*d Wnet=53.3J +13.3*d -DU=0-1/2 *120*d2=-60d2 So 0=60d2+13.3*d+53.3 d=1.06 m

  15. Hints 7.26— Concentrate only on the y-coordinate 7.30— Calculate the work for each path and then sum 7.39— Part a) Each rope carries 1/3 of the weight Part b) Calculate the total work done against gravity Part c) Calculate the total path of the rope 7.42—Part a) Use DK=-DU to calculate Part B) Use W =-DU to calculate where W=F*d*cosine(angle between) 7.46—Part a) Stay on the track the weight must equal the centripetal force Part b) Ignore the path, concentrate on the difference in heights Part c) Use velocity found in part b to compute 7.54—Use the conservation of energy that K+U-Wfriction=E, the total mechanical energy and this E is conserved i.e. K1+U1-Wfriction=U2 in this case 7.66—Use K1+U1=K2+U2-Wfriction 7.62—The skier’s kinetic energy at the bottom can be found by from the potential energy at the top minus the work done by friction. Part b) in the horzintoal, the potential energy is zero so K1=K2-Wfriction-Wair Part c) F*d=DK

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