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10 mWild 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?
At bottom of rope:
Weight of Tarzan=100*9.8 =980 N
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
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
n, normal force (because upside down)
Weight of coaster, mg=1000*9.8=9800
2 kg snap?
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.
2 kg snap?
½ mv2 – 0=53.3
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
7.62—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