A 5.18 kg block located on a horizontal frictionless floor is pulled by a cord t

Question

A 5.18 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F=12.5N at an angle theta=20.0degrees above the horizontal, as shown. What is the magnitude of the normal force acting on the block when the force F is acting on it? Also If, instead, the floor has a coefficient of kinetic friction µk = 0.04, what is the magnitude of the frictional force on the block when the block is moving? Finally, What is the magnitude of the acceleration of the block when friction is being considered?

Explanation / Answer

mg >> down, normal to ground force applied > components F cos 20 >> moves forward F sin 20 >> upwards, counters weight ------------------- Normal force = N = mg - F sin 20 friction force (fc) = uk * N = 0.04*[5.18*9.8 - 12.5*0.342] fc = 1.86 N ---------------- net force (forward) = ma = Fcos 29.5 - fc ma = 9.89 N a = 1.91 m/s^2 = acceleration of block please rate.

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