A body of mass 40,kg resting on a rough horiz | vedclass

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(B) Given: Mass $m = 40\,kg$,coefficient of static friction $\mu_{s} = 0.5$,coefficient of kinetic friction $\mu_{k} = 0.4$,and acceleration due to gr

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(B) Given: Mass $m = 40\,kg$,coefficient of static friction $\mu_{s} = 0.5$,coefficient of kinetic friction $\mu_{k} = 0.4$,and acceleration due to gravity $g = 10\,m/s^2$.The force $P$ is just enough to start the motion,so it must be equal to the limiting friction:$P = f_{s,max} = \mu_{s} N = \mu_{s} mg$.Once the body starts moving,the kinetic friction $f_{k}$ acts on it:$f_{k} = \mu_{k} N = \mu_{k} mg$.The net force $F_{net}$ acting on the body is:$F_{net} = P - f_{k} = \mu_{s} mg - \mu_{k} mg = m(\mu_{s} - \mu_{k})g$.Using Newton's second law,$F_{net} = ma$:$ma = m(\mu_{s} - \mu_{k})g$.Therefore,the acceleration $a$ is:$a = (\mu_{s} - \mu_{k})g$.Substituting the values:$a = (0.5 - 0.4) 	imes 10 = 0.1 	imes 10 = 1\,m/s^2$.

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