A force of 15N is applied at an angle of 30 degrees to the horizontal on a 0.750

Question

A force of 15N is applied at an angle of 30 degrees to the horizontal on a 0.750kg block at rest on a frictionless surface. (a) What is the magnitude of the resulting acceleration of the block? (b) what is the magnitude of the normal force?

Explanation / Answer

Hello, The only force on the block with a horizontal component (in the direction of the acceleration) is: Fcos30° = ma ==> a = (Fcos30°)/m = 15N*cos(30°)/(0.75kg) = 17.32m/s² b). The forces acting on the block along the y-axis are: Fsin(30°) and N upward , and mg downward. From equilibrium along the y-axis (no motion along the y-axis): N + Fsin(30°) = mg N = mg - Fsin(30°) N = 0.75*10 - 15*sin(30°) = 0N

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