Physics Help Forum Two-ish problems(1 pendulum and 1 force problem)

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 Dec 4th 2012, 10:14 PM #1 Junior Member   Join Date: Oct 2012 Posts: 7 Two-ish problems(1 pendulum and 1 force problem) So, I have a test tomorrow, and I just need help with these two problems. I honestly don't know what to do, although, I'm sure if I had the equations I could figure it out. I know the answers for everything so I can check it that way. 1) A 5N force acts on a 2kg object moving with a speed of 4m/s. Over what distance must the force act in order to speed up the object to 6m/s? 2) Tarzan is running at 8.0 m/s top speed and grabs a vine hanging vertically from a tall tree. How high can he swing up? Does the length of the vine affect your answer? Also, I have one where I got the right answer, but I'm not sure I arrived at it the right way. The problem is... A block slides down a frictionless incline and the slides across a floor with a coefficient of kinetic friction of .18. How much work is done by friction on the block? Har far does the block slide before coming to rest? I got my distance, 14m, now, I know the work done is -49J. Is that because mgh=.5mv^2+f.x.cos(theta) therefore mgh is 0 because the block is no longer on the incline so the height is 0, and f =uk*fn, x=14, f*x=49?
 Dec 5th 2012, 07:30 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,354 For the first problem you can use work and energy principals: the work done by the force 'F' acting over a distance 'd' equals the change in kinetic energy of the object: For the second problem use energy principals again: change in potential energy + change in kinetic energy = 0. The potential energy he gains by attaining height 'h' is equal to mgh. His starting kinetic energy is (1/2)mv^2 and his KE at the top of the arc is 0, because at that point he is motionless. Now you can solve for 'h'. For the third question I'm not following you, but in general the work done by the force of friction as the object slides horizontally must equal the initial KE when it first starts to slide on the floor, and that KE in turn must equal the loss of PE that occurred as it slid down the incline. So friction force = umg, and work by friction = umgd, which equals mgh. Hence d = h/u. Last edited by ChipB; Dec 5th 2012 at 10:15 AM.

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