(B) Given, tension in the string $(T) = 27 \,N$.Mass of the hanging block $(m) = 3 \,kg$.Let the acceleration of the system be $a$.For the hanging blo

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(B) Given, tension in the string $(T) = 27 \,N$.Mass of the hanging block $(m) = 3 \,kg$.Let the acceleration of the system be $a$.For the hanging block of mass $3 \,kg$, the equation of motion is:$m g - T = m a$$3 \times 10 - 27 = 3 a$$30 - 27 = 3 a$$3 = 3 a$$a = 1 \,m/s^2$.For the body of mass $M = 20 \,kg$ moving on the horizontal surface, the equation of motion is:$T - f_k = M a$where $f_k = \mu M g$ is the kinetic friction force.$27 - \mu \times 20 \times 10 = 20 \times 1$$27 - 200 \mu = 20$$200 \mu = 27 - 20$$200 \mu = 7$$\mu = \frac{7}{200} = 0.035$.Therefore, the coefficient of kinetic friction is $0.035$.

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 mass $20 \,kg$ is moving on a rough horizontal plane. $A$ block of mass $3 \,kg$ is connected to the $20 \,kg$ mass by a string of negligible mass through a smooth pulley as shown in the figure. The tension in the string is $27 \,N$. The coefficient of kinetic friction between the heavier mass and the surface is $\left(g=10 \,m/s^2\right)$.

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A
$0.025$
B
$0.035$
C
$0.35$
D
$0.25$

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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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$A$ body of mass $1 \,kg$ is moving with a velocity $10 \,ms^{-1}$ on a horizontal rough surface having a coefficient of kinetic friction $0.4$. If the constant force is removed, the body comes to rest in a time (Acceleration due to gravity $g = 10 \,ms^{-2}$). (in $\,s$)

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