A 7.0N force acts on a 1.5 kg mass to give it an acceleration of 4.0 m/s2, south. The force of friction, including direction, is
Fnet=ma→7−Ff=1.5⋅4→Ff=0F_{net}=ma\to 7-F_f=1.5\cdot4\to F_f=0Fnet=ma→7−Ff=1.5⋅4→Ff=0
Based on the condition of the problem Ff=0F_f=0Ff=0 . Answer
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