A block of mass M = 100 , kg is pulled by a f | vedclass

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(C) For the block,the forces are balanced as follows:$T \cos 37^{\circ} = f$$N + T \sin 37^{\circ} = Mg$Since $f = \mu N$,we have $T \cos 37^{\circ} =

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$\frac{g}{3}$

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(C) For the block,the forces are balanced as follows:$T \cos 37^{\circ} = f$$N + T \sin 37^{\circ} = Mg$Since $f = \mu N$,we have $T \cos 37^{\circ} = \mu (Mg - T \sin 37^{\circ})$.Substituting the values $M = 100 \, kg$,$\mu = 1/3$,$\cos 37^{\circ} = 4/5$,and $\sin 37^{\circ} = 3/5$:$T(4/5) = (1/3)(100g - T(3/5))$$12T/5 = 100g - 3T/5$$15T/5 = 100g \implies 3T = 100g \implies T = 100g/3$.For the person,the upward force is $T = 100g/3$ and the downward force is $mg = 25g$.The equation of motion is $T - mg = ma_{\max}$.$100g/3 - 25g = 25a_{\max}$$(100g - 75g)/3 = 25a_{\max}$$25g/3 = 25a_{\max}$$a_{\max} = g/3$.

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