(D) Given: Weight $W = 64 \,N$, coefficient of static friction $\mu_s = 0.6$, coefficient of kinetic (dynamic) friction $\mu_k = 0.4$. The force appli

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(D) Given: Weight $W = 64 \,N$, coefficient of static friction $\mu_s = 0.6$, coefficient of kinetic (dynamic) friction $\mu_k = 0.4$. The force applied to start the motion is equal to the limiting friction: $F = f_{s,max} = \mu_s N = \mu_s W$. Since $W = mg$, we have $F = 0.6 \times mg$. Once the body starts moving, the kinetic friction acting on it is $f_k = \mu_k N = \mu_k mg = 0.4 \times mg$. The net force acting on the body is $F_{net} = F - f_k = 0.6 mg - 0.4 mg = 0.2 mg$. According to Newton's second law, $F_{net} = ma$. Therefore, $ma = 0.2 mg$, which gives $a = 0.2 g$.

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

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$A$ body of weight $64 \,N$ is pushed with just enough force to start it moving across a horizontal floor and the same force continues to act afterwards. If the coefficients of static and dynamic friction are $0.6$ and $0.4$ respectively, the acceleration of the body is (Acceleration due to gravity $= g$)

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A
$\frac{g}{6.4}$
B
$0.64 g$
C
$\frac{g}{32}$
D
$0.2 \,g$

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Class 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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