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Question

A force of 250 N pushes a 50 kg box along a horizontal surface. The force is directed 30o below the horizontal. What is the acceleration of the box if the coefficient of kinetic friction between the box and the surface is 0.30? (Since the possible answers given are written to two digits of accuracy, use g = 9.8 m/s2.)

Answer 1

The box has horizontal acceleration given by

 

a_x = F_xnet/m = (Fcosθ - μN)/m

 

F_xnet = net horizontal force

F = applied force = 250 N

m = 50 kg

N = upward normal force

 

N = mg + Fsinθ  [upward normal balances downward weight and pull]

 

a_x = [Fcosθ - μ(mg + Fsinθ)] / m

 

a_x = [250cos30° - 0.30(50*9.8 + 250sin30°)] / 50 m/s²

 

a_x ≅ [216.51 - 0.30(615)] / 50 m/s² ≅ 0.640 m/s²