A block of mass 40 ,kg slides over a surface, | vedclass

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(D) Let $M = 40 \,kg$ be the mass of the block on the surface and $m = 4 \,kg$ be the suspended mass. Let $a$ be the acceleration of the system and $T

Vedclass · Accepted Answer

$8/11$

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(D) Let $M = 40 \,kg$ be the mass of the block on the surface and $m = 4 \,kg$ be the suspended mass. Let $a$ be the acceleration of the system and $T$ be the tension in the string.For the suspended mass $m$,the equation of motion is:$mg - T = ma$$4(10) - T = 4a \implies 40 - T = 4a$  --- $(1)$For the block $M$ on the surface,the kinetic friction force is $f_k = \mu_k N = \mu_k Mg = 0.02 	imes 40 	imes 10 = 8 \,N$.The equation of motion for the block is:$T - f_k = Ma$$T - 8 = 40a$ --- $(2)$Adding equations $(1)$ and $(2)$:$(40 - T) + (T - 8) = 4a + 40a$$32 = 44a$$a = \frac{32}{44} = \frac{8}{11} \,m/s^2$.

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