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5 November, 02:51

A 3kg iron block rests on a floor of an unknown material. Through experimentation, the coefficient of kinetic friction between them is 0.85. What applied force is required to accelerate the block at 0.15 m/s2?

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  1. 5 November, 06:06
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    F=ma by newtons 2nd law where F = the net force acting on the system. to have acceleration there must enough force applied to overcome the frictional force which always acts in the opposite direction as the applied force. with this in mind we can set up the equation with the net total force on the system being

    Fa (the applied force) - mgf where m=mass, g=gravity, f=friction (because the opposing frictional force is just the coefficient of friction * the normal force and the normal force is just mg. so the with Fnet = ma we have

    Fapplied-mgf=ma this gives us

    F (applied) - (3kg*9.81m/s^2*.85) = 3kg*.15m/s^2 from here simply solve for F which gives F = 3kg*.15m/s^2+3kg*9.81m/s^2*.85 which gives F=25.5kg*m/s^2 I left the units in so you can see they lead to kg*m/s^2 which is equal to Newtons in terms of units so then F=25.5N
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