Document 4-1 PHYSICS- \"problems [Sheet #11 In the following problems, the coeff

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Document 4-1 PHYSICS- "problems [Sheet #11 In the following problems, the coefficient of fiction is given by ·E , where Fs is the normal force (simply the object's weight): also, use 9.8 m's or 32 fts where appropriate for the accel eration due to gravity 1.) What F,wil exist when sliding a 5 kg book along a table surface where the coefficient of friction (H. pro nounced "mu) between table and book is 2? 2.) What mass is being pulled along a surface if a coeffcient of fricton of5 between the mass and surface causes a F of 20 N? 3.) A1.5 slug mass is pulled with a spring scale at a slow and constant speed. If the spring scale reads 6 lb while puling the mass, then what was the between the surface and mass ? a-- 1 5 slug 4.) A70kgbox is slid on a floor by a 400 N force. The u between the box and floor is 5. Find the acceleration of the box on the floor. 70kg -400 N 5) 05-lf an inner tube (of 205 kg mass) were The between an inflated nner ute and a patch of ice is moving at 5 m/s as it began to slide on the ice, find: (a) the F, in N) between the ice and inner tube. (b) the acceleration rate (in m/s) of the inner tube (c) the time (in seconds) necessary for the inner tube to come to a compiete stop (d) the dstance (in meters) necessary for the inner tube to come to a complete stop. p. 1

Explanation / Answer

1] friction Ff = uFN = umg = 0.2*0.5*9.8 = 0.98 N answer

2] From above m =  Ff / ug = 20/[0.35*9.8] = 5.831 kg

3] Here acceleration a = 0

So friction = tension = 6lbf and mass = 1.5 slug = 1.5*32.174 = 48.26 lb

coefficient of friction = 6/48.26 = 0.124

4] acceleration a = Fnet/m = [400- 0.5*70*9.8]/70

= 0.814 m/s^2 answer and friction = umg = 0.5*70*9.8= 343 N

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