- #1
physaru86
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A 1.2 kg block is resting on horizontal surface. The coefficient of static friction between block and
surface is 0.5. What is the magnitude and direction of force of friction on block when magnitude of external force acting of block in the horizontal direction is 9.8 N fs = μs*N, N = mg
m = 1.2 kg, μs = 0.5, g = 9.8
fs = 0.5*1.2*9.8 N = 5.88 N
surface is 0.5. What is the magnitude and direction of force of friction on block when magnitude of external force acting of block in the horizontal direction is 9.8 N fs = μs*N, N = mg
m = 1.2 kg, μs = 0.5, g = 9.8
fs = 0.5*1.2*9.8 N = 5.88 N