Recent content by CandyApples

Yes it is now much easier, thank you so much.

Homework Statement Tarzan mass M swings from a vine from rest at height H. At the bottom of the arc he picks up Jane of mass m in a perfectly inelastic collision. They reach a vertical height h. How far vertically above h is H?Homework Equations Pi = Pf R = \Sigmamr/\Sigmam The Attempt at a...

Ohhhh and since Py initial is zero and the mass/velcoity of car 2 is known as well as final v for car 1 this becomes much easier. Funny how problems become less challenging when I utilize both dimensions.

Homework Statement Two cars collide as shown. Car s1 is initially moving along the x-axis at speed v1 and car 2 is stationary. The final speed of car one is 10 m/s and goes 30 degrees north of x-axis. The mass of car 2 is 2000kg and goes 6 m/s at 44 degrees south of x-axis. What is the mass and...

Homework Statement A bungee jumper jumps with a mass of 61kg jumps off of a bridge. The massless bungee cord is 25m long and beyond that length acts as a spring. The jumper reaches the lowest point 42.9m below the bridge. Find the force constant which characterizes the bungee cord. Homework...

Ok so now there is a part two to this problem. I figured out Part I as follows: Fs = .5 (147) 73.5 = 15*a a = 4.9m/s/s Fk = 392*.4 Fk = 156.8 F-156.8 = ma F = 40(4.9)+157 F = 353N Now it asks if F determined in Part I is doubled (706N) what will be the acceleration of each block...

Homework Statement A man is pulling two suitcases, one on top of the other. Bottom case mass=25kg and top mass = 15kg. The coefficient of friction between the two cases and between the bottom case and the floor can both be represented by Uk = .4 and Us =.5. Find the maximum acceleration of the...

Homework Statement A ride at a carnival rotates about a vertical axis. When spinning fast enough the bottom is dropped and the rider sticks to the side. Find the minimum tangential speed so that the rider does not fall. R=radius of circle, u= coefficient of static friction and g represents...

My apologies, I overcomplicated this problem. Upon drawing a diagram it is obvious Fx = T2cos(50) - 150cos(40). Fy = T2sin50 + 150sin40. Apply second law to Fx and force comes out nicely to 178.8N, plug that into Fy and it is in fact 23.8kg. Thanks for the help :).

Homework Statement A large chandelier is supported by two ropes. Rope 1 makes a 40 degree angle with the ceiling and has a tension of 150N. Rope two forms a 50 degree angle with the ceiling. What is the tension in rope 2 and what is the mass of the chandelier given the chandelier...

oh my, it was 235. The new a is 2 and hence the tension is 220. Got too enthused that I solved the problem and made a silly mistake. Thank you very much for the assistance and thank you for saving me some embarrassment :).

the person on the slope would have the forces: \SigmaFx = T \SigmaFy = 0 Can I solve using T = 110(a) in my x force component for the other person, seeing as tension is constant throughout the entire rope? Doing so gives: T = 110a F = -t + mgsin(22.56) 125a = -110a +125(9.8)(sin(22.56)) 225a =...

Homework Statement Two people are trying to climb an iced-over mountain. They are connected by a rope which passes smoothly over a rock where the terrain bends (acting as a pulley). One climber (110kg) is on top on a horizontal surface while the other (125kg) is on a 22.56 degree incline...

Homework Statement A model rocket is launched at 50m/s, 35 degrees above horizontal. What is the horizontal displacement when its velocity vector is at 25 degrees. Homework Equations Kinematic equations, the relationship between sin cos and tan. The Attempt at a Solution tan(25)=...

oh should it be a +.9 since at t=0 he is .9 above where he will end up?