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HW-Sec 1-Prob 21

# HW-Sec 1-Prob 21 - calculated by force =(400J(0.05m =...

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Robb Page 08-Mar-2010 Physics 140 Professor Stoddard Chapter 1 – Problem 21 Ques. The wedge in problem 20 acts like a ramp, slowly splitting the wood as it enters the log. The work you do on the wedge, pushing into the log, is the work it does on the wood, separating its two halves. If the two halves of the log only separate by a distance of 0.05 m while the wedge travels 0.2 m into the log, how much force is the wedge exerting on the two halves of the log to rip them apart? Since this is a continuation type of problem, it becomes a rather simple reformulation of the answer from question 20. Since we can calculate work as: work = force • distance, we can flip the data in order to calculate the force, therefore: force = work / distance Using the answer from problem 20, we have 400J of work. Distance, in this instance is the amount of separation of the log NOT the movement of the wedge, and is provided as 0.05m. Therefore, the force on the object is
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Unformatted text preview: calculated by: force = (400J) / (0.05m) = 8000N (Again, N means “measured in Newtons”). The wedge is acting as simple machine, by extension of the force from the blow of the mallet being transferred to the contact point of the wedge-to-wood. As the book points out, mechanical advantage “allows you to do work that would require unrealistically large force, but instead can achieve results exerting much smaller force for a much longer distance,” (pg. 37). Therefore, the energy applied by each blow, transfers ample concentrated force into the wedge, and the wedge, moving at its 0.2m per blow, is able to maximize that energy-to-force transference (displacement), overcoming both the friction and material resistance of the log resulting, eventually, in two halves. F net F 1 Split-per blow (displacement 0.05m) Kinetic energy F 2 Wedge-per blow (movement 0.2m)...
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