A force of 65.9 N is required to start a 7.0 kg box movingacross a horizontal co

Question

A force of 65.9 N is required to start a 7.0 kg box movingacross a horizontal concrete floor. *What is the coefficient of static friction between the boxand the floor? *If 65.9 N force continues, the box accelerates at 0.50m/s^2. What is the coefficient of kinetic friction? A force of 65.9 N is required to start a 7.0 kg box movingacross a horizontal concrete floor. *What is the coefficient of static friction between the boxand the floor? *If 65.9 N force continues, the box accelerates at 0.50m/s^2. What is the coefficient of kinetic friction?

Explanation / Answer

Along the horizontal direction              F - fs = 0              F = smg               s = F / mg                        = 65.9 / 7 * 9.8                        = 0.960 If the box is accelerating then               F - fs = ma               k = (F - ma) / mg                    = (65.9 - 7*0.5) / 7*9.8                    = 0.909

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