A 2.6kg box is pulled across a flat frictionless table by a 22N force. what is the acceleration of the box? what is the acceleration with the kinetic coefficient of friction is 0.3?

Question

anonymous · Answer

F-R=ma, since it is frictionless,mthen R is 0 and a=F/m= 22/2.6=8.46m/s^2.
N or normal contact force is the same as mg or 2.6*10= 26N.
Ff= €N, where € is coefficient of friction, then Ff=0.3*2.6=0.78N.
Recoiling the formula F-R=ma, making a the subject is (F-R)/m= (22-0.78)/2.6=8.2m/s^2.

michele_laino · Answer

when a friction force also acting on our box, we have this situation:
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