A force is applied to an initially stationary block of mass 1.75 kg that sits on a horizontal floor as shown. The 53.5 N force is applied at θ = 32° angle. The coefficients of friction between the floor and the block are μs = 0.593 and μk = 0.343. What is the acceleration of the block?

Answer 1

m*g = 1.75 * 9.8 = 17.15 N. = Wt. of block.

Fn = 17.15 + 53.5*sin32 = 45.5 N. = Normal = Force perpendicular to the floor.

Fk = u*Fn = 0.343 * 45.5 = 15.61 N. = Force of kinetic friction.

a = (Fx-Fk)/m = (53.5*Cos32-15.61)/1.75 = 17 m/s^2.

Answer 2

Henry’s solution is wrong because it fails to realize that u_s is a great enough coefficient of static friction to prevent the block from accelerating at all. Acceleration is zero