(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

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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$$4(10) - T = 4a \implies 40 - T = 4a$ --- $(1)$For the block $M$ on the surface,the kinetic friction force is $f_k = \mu_k N = \mu_k Mg = 0.02 \times 40 \times 10 = 8 \,N$.The equation of motion for the block is:$T - f_k = Ma$$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$.

Class 11 Physics Newton's Laws of Motion and Friction Kinetic Friction and Motion on Rough Horizontal Surface

Medium

$A$ block of mass $40 \,kg$ slides over a surface,when a mass of $4 \,kg$ is suspended through an inextensible massless string passing over a frictionless pulley as shown below. The coefficient of kinetic friction between the surface and the block is $0.02$. The acceleration of the block is ............ $m/s^2$. (Given $g = 10 \,m/s^2$.)

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A
$1$
B
$1/5$
C
$4/5$
D
$8/11$

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Newton's Laws of Motion and Friction

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Kinetic Friction and Motion on Rough Horizontal Surface

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