text: physics question? what force must act horzontally on a 20.0 kg mass moving at a constant speed of 4.00m/s on a rough table of coefficient of kinetic friction of 0.300? if the force is removed when will the body come to rest? where will it come to rest?  show all work and formulas used with correct answer to cecieve best answer. thanks for the help! fweigh=mg fweigh= 20kg(9.8) fweigh= 196n fnormal is also 196n  friction= mu(fnormal) friction= (.3)(196n) friction= 58.8n  everytime you see the word "constant speed" or "motionless" that mean all the forces are balanced. so the friction="horizonal force" so 58.8 is the horizonal force that keep the object moving at a constant speed.  we have to find acceleration. a=fnet/m  ffriction is the fnet because the horizonal force is romoved a=friction/m a= 58.8n/20kg a= 2.94m/s^2 (should be -2.94m/s^2 because it is slowing down)  vf^2=2ad+vi^2 (0m/s)^2=2(-2.94m/s^2)d+(4m/s)^2 -16m^2/s^2= (-5.88m/s^2)d d= 2.72m