A block of mass 5 kg is kept on a horizontal floor having coefficient of friction 0.09. Two mutually perpendicular horizontal forces of 3 N and 4 N act on this block. The acceleration of the block is x m/s2. Find value of x. (g = 10m/s2)
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answer is 0.1.
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Detailed Solution
Net external force F=(4)2+(3)2=5NMaximum friction tmax=μmg=(0.09)(5)(10)=4.5NSince F>tmax, block will not move with a n acceleration, a=F−fmaxm=5−4.55=0.1m/s2