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A block of mass 2 kg is on a horizontal surface the cofficient of Static and kinetic friction are 0.6 and 0.2 the minimum horizontal force required to start the motion is applied and if it is continued the velocity acquired by the body at the end of the two seconds is( g= 10 ms^2)
M=20kgmuMg = 2 * 10 * 0.6 (static Smax)= 12N force is applied to the blockmuMg = 2 * 10 * 0.2 (kinetic)=4NFnet = F-fr = 12 - 4 = 8NF/m=8/2= 4 m/s2#use v=u+at u= 0, a=4 , t= 2 Final velocity will be 8m/s RegardsArun
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