Two blocks of masses `m_(1) = 10 kg` and `m_(2) = 20 kg` are connected by a spring of stiffness k = 200 N/m. The coefficient of friction between the blocks and the fixed horizontal surface is `mu = 0.1`. Find the minimum constant horizontal force F (in Newton) to be applied to m1 in order to slide the mass `m_(2)`. (Take `g = 10 m//s^(2)`)

Explanation

WF+WSp+W fric=`DeltaK`
`rArr Fx-1/2Kx^(2)- mu m_(1)g x=0` and `Kx=mu m_(2)g`
`rArr F-1/2 mu m_(2)g-mu m_(1)g=0 rArr F=mu m_(1)g+(mu m_(2)g)/2=0.1xx10xx10+(0.1xx20xx10)/2 =20 N`