A person is applying a horizontal force in trying to push a 25-kgblock up a fric

Question

A person is applying a horizontal force in trying to push a 25-kgblock up a frictionless plane inclined at an angle of 15 degrees.A.) calculate the force needed just to keep the block inequilibrium. B.) Suppose that she applies three times that force.What will be the acceleration of the block?

Explanation / Answer

A) let the force be F, here a =0 Fcos(15) - mgsin(15) = ma = 0 =>F = mgtan(15) = 25kg*9.8m/s2 *tan(15) = 65.65N 65.65 N force us needed to keep equilibrium B) now F = 65.65*3 = 196.95N Fcos(15) - mgsin(15) = ma =>a = [Fcos(15) - mgsin(15) ]/ m = [196.95N*cos(15) -25kg*9.8m/s2 sin(15)]/25kg = 5.07m/s2

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