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Exercise · Q15

Q.A trunk of mass 20 kg20\ \text{kg} is dragged at constant velocity across a horizontal floor by a horizontal force of 60 N60\ \text{N}. Calculate the coefficient of kinetic friction between the trunk and the floor. (Take g=9.8 m/s2g = 9.8\ \text{m/s}^2.)

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Given: mass m=20 kgm = 20\ \text{kg}, applied force F=60 NF = 60\ \text{N}, constant velocity (so zero acceleration), g=9.8 m/s2g = 9.8\ \text{m/s}^2.\n\nSince the trunk moves at constant velocity, its acceleration is zero, so by Newton's second law the net force on it must also be zero -- meaning the applied force is exactly balanced by kinetic friction: F=fk=μkN=μkmgF = f_k = \mu_k N = \mu_k mg\n\nOn the horizontal floor, $N = mg = 20 \times 9.8 = 196\ \text{N} …

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