A 12 kg box at rest on the floor requires a horizontal force of 47 N to start moving it. Once it starts moving, the same horizontal force makes the box accelerate at 1.1m/s^2. What is the coefficient of kinetic friction?

Question

anonymous · Answer

Am i on the right track? I calculated the Fnet =ma =(12kg)(1.1m/s^2)=13.2N. Then i calculated to find Ff, Fnet=Fa+Ff, 13.2N=47N+Ff , Ff=33.8N. Then i found Fn=mg=(12)(9.8)=117.6n. Lastly i found the coefficient of kinetic friction: COF=Fk/Fn = 33.8n/ 117.6n = 0.29