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(A) Given: Mass $M = 5\,kg$,coefficient of friction $\mu_k = 0.2$,applied force $F = 40\,N$,angle $	heta = 30^\circ$,and $g = 10\,m/s^2$.The normal r

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$5.73$

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(A) Given: Mass $M = 5\,kg$,coefficient of friction $\mu_k = 0.2$,applied force $F = 40\,N$,angle $	heta = 30^\circ$,and $g = 10\,m/s^2$.The normal reaction $R$ on the block is given by $R = Mg - F\sin 30^\circ$.$R = (5 	imes 10) - (40 	imes \sin 30^\circ) = 50 - (40 	imes 0.5) = 50 - 20 = 30\,N$.The kinetic friction force $f_k$ is given by $f_k = \mu_k R$.$f_k = 0.2 	imes 30 = 6\,N$.The net force acting on the block in the horizontal direction is $F_{net} = F\cos 30^\circ - f_k$.$F_{net} = (40 	imes \cos 30^\circ) - 6 = (40 	imes \frac{\sqrt{3}}{2}) - 6 = 20\sqrt{3} - 6$.Using $\sqrt{3} \approx 1.732$,$F_{net} = 20(1.732) - 6 = 34.64 - 6 = 28.64\,N$.The acceleration $a$ of the block is $a = \frac{F_{net}}{M} = \frac{28.64}{5} = 5.728\,m/s^2 \approx 5.73\,m/s^2$.

$A$ block of mass $M = 5\,kg$ is resting on a rough horizontal surface for which the coefficient of friction is $0.2$. When a force $F = 40\,N$ is applied at an angle of $30^\circ$ with the horizontal,the acceleration of the block will be ........ $m/s^2$ $(g = 10\,m/s^2)$.

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