a block mass 9 kg is at rest on inclined plane sloped at angletheta form horiz.

Question

a block mass 9 kg is at rest on inclined plane sloped at angletheta form horiz. Block is located 5 m from bottof ofplan. What is frictional for for theta = 20 degrees If angle increased slowly block starts sliding at angle 30degrees. What is coefficient of static friction? If block slides to bottom in 2 s ( theta - 30) what iscoefficent of kinetic fricktion a block mass 9 kg is at rest on inclined plane sloped at angletheta form horiz. Block is located 5 m from bottof ofplan. What is frictional for for theta = 20 degrees If angle increased slowly block starts sliding at angle 30degrees. What is coefficient of static friction? If block slides to bottom in 2 s ( theta - 30) what iscoefficent of kinetic fricktion

Explanation / Answer

         =30.2N
In the second case when the block just start sliding, thedownward component of its weight(mgsin) is equal to theupward static friction force (mgcos) therefore we have mgsin = mgcos thrn coefficient of ststic friction =tan          =tan(30o)    = 0.577
In this case the acceleration of teh block is gsin- gcos = (9.8m/s2)(sin20o- cos20o) =  (9.8m/s2)(0.342 -0.93969) Now using the equation x =voxt+(1/2)axt2 As the initial velocity is zero therefore a = 2x/t2    = (2)(5m)/(2s)2 = 2.5m/s2
therefore (9.8m/s2)(0.342 - 0.93969)= 2.5m/s2 0.342 - 0.93969 =0.255 or coefficient of kinetic friction k =0.09258

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