The coefficients of static and kinetic fricti | vedclass

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(D) The minimum force required to start motion is equal to the limiting static friction: $F_{lim} = \mu_s N = \mu_s mg$.Given $F_{lim} = 64 \, N$ and

Vedclass · Accepted Answer

$0.2 \, g$

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(D) The minimum force required to start motion is equal to the limiting static friction: $F_{lim} = \mu_s N = \mu_s mg$.Given $F_{lim} = 64 \, N$ and $\mu_s = 0.6$,we have $64 = 0.6 	imes m 	imes g$,which gives the mass $m = \frac{64}{0.6g}$.Once the block is in motion,the kinetic friction acting on it is $f_k = \mu_k N = \mu_k mg$.Substituting the values: $f_k = 0.4 	imes m 	imes g = 0.4 	imes \left( \frac{64}{0.6g} ight) 	imes g = 0.4 	imes \frac{64}{0.6} = \frac{25.6}{0.6} = \frac{128}{3} \approx 42.67 \, N$.The net force acting on the block is $F_{net} = F_{applied} - f_k = 64 - \frac{128}{3} = \frac{192 - 128}{3} = \frac{64}{3} \, N$.The acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{64/3}{64/(0.6g)} = \frac{64}{3} 	imes \frac{0.6g}{64} = \frac{0.6g}{3} = 0.2g$.

The coefficients of static and kinetic friction for a block on a rough surface are $0.6$ and $0.4$ respectively. The minimum force required to set the block in motion is $64 \, N$. If this force continues to be applied after the block starts moving,what will be the acceleration of the block?

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