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PHY121 Ch 6-9 Exam Name You are a member of an alpine rescue team and must get a 3.00 kg box of supplies up a 30° incline so that it reaches a stranded skier who is a vertical distance 4.00 m above the bottom of the incline. There is some friction present, µ = 0.100. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. To aid in the ascent, you attach one of your bottle rockets to the back side of the box of supplies, as shown, which exerts a force of 10.0 N. Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier. TEbottom = TEtop [KEpush + PEbottom] + FR-in dir d – Ff d = KEtop + PEtop KEpush + PEbottom + FR-in dir d – Ff d = 0 + mghtop ½3v2 + 0 + 10 cos30° 8 µ FN 8 = m g htop 2 ½3v + 0 + 69.3 µ FN 8 = 3 10 (4) 5/5 3/3 5/5 5/5 5/5 5/5 Note, please account for energy FR d in the initial energy, similar but opposite to the non-conservative force, Friction contributed by the bottle rocket, through a distance, 5/5 2 1.5v + 69.3 - (0.1)(10sin30 + mgcos30) 8 v = 7.1 m/s = Ff d. sin θ = h / d sin 30 = 4 / d d = 8 meters 1/1 120 J A 50.0 kg roller-coaster car is released from rest and slides along a frictionless track. At ground level, the car encounters a loop of radius R = 10 m. If the track exerts a normal force of 625 N on the car at the top of the loop, what is the initial height of the car? Both mg and the normal force are toward the center, thus additive. 2/2 8/8 2/2 7/7 mg + 700 = mv2 / r 50(10) + 625 = 50 v2 / 10 50(10) + 625 = 50 v2 / 10 v =15 m/s 4/4 8/8 2/2 7/7 2 mg + 200 = mv / r 50(10) + 200 = 50 v2 / R = mghtrack + ½mvtrack 500(htrack) + 0 htrack = 31.25 m A 50.0 kg roller-coaster car is moving 4 m/s at h = 30 meters on the top of the track and slides along a frictionless track. At ground level, the car encounters a loop of radius, R. If the track exerts a normal force of 200 N on the car at the top of the loop, what is R? 2/2 TEtoptrack 2 TEtoploop 2/2 = mghloop + ½mvloop2 = 500(2R) + ½50 152 OR for 80% max Both mg and the normal force are toward the center, thus additive. 4/4 8/8 TEtoptrack A 50.0 kg roller-coaster car is moving 10 m/s at h = 30 meters on the top of the track and slides along a frictionless track. At ground level, the car encounters a loop of radius, R = 10 m. What is the normal force exerted by the track on the car at the top of the loop? = TEtoploop 2/2 mghtrack + ½mvtrack 2/2 2 = mghloop 4/4 + ½ m vloop2 14 R = vtoploop2 4/4 TEtoptrack = 2/2 mghtrack + ½mvtrack2 500(30) + ½50 42 15400 R = 11.4 meters 2/2 TEtoploop 500(30) + ½50 102 vloop = 17.3 m/s = 500(2R) + ½50 vloop2 2/2 6/6 2/2 4/4 mg + FN = m v2 / r 50(10) + FN = 50 17.32 / 10 FN = 1000 N 6/6 = mghloop + ½mvloop2 = 500(2R) + ½50 14R = 1350R A 0.500 kg block rests on a frictionless table and is attached to a horizontal spring (k = 30 N/m) that is uncompressed. A 0.100 kg wad of putty is launched horizontally on the frictionless table by another identical spring which is compressed by 10 cm. The wad impacts the blocks and sticks. How far does the putty-block system compress the original spring? Workspring = ΔK po = pf Workspring = ΔK 5/5 4/4 2 7/7 2 ½kx =½mv ½30(.1)2 = ½(.1)vwad2 vwad = 1.73 m/s 7/7 5/5 5/5 2 ½ k x = ½(mblock + mwad) vf2 ½30x2 = ½(mblock + mwad) vf2 x = 4.08 cm mvblock + mvwad = (mblock + mwad) vf 0 + .1 (1.73) = (0.5 + 0.1) vf vf = 0.289 m/s The collision between a sledge hammer and a nail for a rail tie can be considered to be approximately elastic. Calculate the kinetic energy acquired by a 500-g nail when it is struck by a 8 kg hammer moving with an initial speed of 5.0 m/s. 8/8 8/8 mh vh + mn vn = mh vhf + mn vnf 8 (5) + ½ (0) = 8vhf + ½vnf 80 - 16vhf = vnf 4/4 4/4 2 ½mn vn2 2 4/4 = ½ mh vhf2 + ½mn = ½ 8 vhf2 + ½ ½ = vnf2 = (80 - 16 vhf)2 ½mh vh + ½8 52 + ½ ½ 0 400 - 16 vhf2 400 - 16 vhf2 Vhf = 4.41 m/s; Vnf = 9.44 m/s 5/5 2 vnf vnf2 Ko = 0 Kf = ½ m v2 Kf = ½ ½ 9.442 Kf = 22.3 J