A force is applied to an initially stationary block of mass 2.05 kg that sits on

Question

A force is applied to an initially stationary block of mass 2.05 kg that sits on a horizontal floor as shown. The 60.9 N force is applied at ? = 38 angle. The coefficients of friction between the floor and the block are ?s = 0.681 and ?k = 0.401. What is the acceleration of the block?

I am having touble coming up with the correct formulas. An explanation of each step would be helpful please.

A force is applied to an initially stationary block of mass 2.05 kg that sits on a horizontal floor as shown. The 60.9 N force is applied at ? = 38 angle. The coefficients of friction between the floor and the block are ?s = 0.681 and ?k = 0.401. What is the acceleration of the block? I am having touble coming up with the correct formulas. An explanation of each step would be helpful please.

Explanation / Answer

sum forces in the y

N - F cos theta - m g = 0

N = 2.05*9.81 + 60.9*cos(38 degrees)= 68.1 N

so F static max = 0.681*68.1= 46.4 J

force in the x = 60.9*sin(38 degrees)= 37.5 N

since the forces can't overcome the static friction, a = 0

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