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View History Bookmarks People Window Hel ssment And × usuHome- ps://edugen.wileyplus.com/edugen/student/mainfr.uni d Mail-jbrya45@isu.edu x : My Home xWileyPLUS . ent FULL SCREEN PRINTER VERSION BACK Chapter 10, Problem 14 A 14.1-ka block rests on a horizontal table and is attached to one end of a massless, horizontal spring, By pulling horizontally on the other causes the block to accelerate uniformly and reach a speed of 4.93 m/s in 0.645 s. In the process, t rtdoare constant speed of 4.93 m/s, during which time the spring is stretched by only 0.0891 m. Find ( between the block and the table. the spring is stretched by 0.186 m. The block is then pulled at a (a) the spring constant of the spring and (b) the coefficlent of kinetic friction (a) NumberT Units (b) Number Units Question Attempts: Unlimited SAVE FOR LATER SUBMKET ANSWER Version 4.24.24 CY PolisI £ 2000-2017 Jobn Weey & 5ons. Inc All Rights Reserved. A Division of bn wiley & Sons, Inc.

Explanation / Answer

Block's acceleration = a = 4.93 m/s / 0.645s

a = 7.643 m/s2

Net accelerating force on the block is
fa = m*a

fa = 14.1*7.643 N

fa = 107.772 N

ff = 0.0891m * k

total stretching force is
fa + ff = k*x

107.772 + 0.0891*k = 0.186*k
k = 107.772 / (0.186 - 0.0891) N/m

k = 1112.2 N/m

k = 1.11 x 103 N/m

Part b)

ff = k*m*g

ff = 0.0891*1786 N

ff = 99.09N

k = 89.3 / (14.1*9.8)

k = 0.717

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