(B) Given: Mass $m = 10 \,kg$, Coefficient of friction $\mu = 0.3$, Applied force $F = 50 \,N$, Acceleration due to gravity $g = 10 \,ms^{-2}$.First,

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(B) Given: Mass $m = 10 \,kg$, Coefficient of friction $\mu = 0.3$, Applied force $F = 50 \,N$, Acceleration due to gravity $g = 10 \,ms^{-2}$.First, calculate the limiting frictional force $f_l = \mu mg = 0.3 \times 10 \times 10 = 30 \,N$.Since the applied force $F = 50 \,N$ is greater than the limiting frictional force $f_l = 30 \,N$, the body will move.The net force acting on the body is $F_{net} = F - f_l = 50 \,N - 30 \,N = 20 \,N$.Using Newton's second law, $F_{net} = ma$, we get $20 = 10 \times a$.Therefore, the acceleration $a = 2 \,ms^{-2}$.

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 $10 \,kg$ is kept on a rough horizontal surface with a coefficient of friction of $0.3$. If a horizontal force of $50 \,N$ is applied to the body, then the acceleration of the body is (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,ms^{-2}$)

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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 $10 \,kg$ is kept on a rough horizontal surface with a coefficient of friction of $0.3$. If a horizontal force of $50 \,N$ is applied to the body, then the acceleration of the body is (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,ms^{-2}$)

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Meena applies the front brakes while riding her bicycle along a flat road. The force that slows her bicycle is provided by the

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When the engine is switched off,a vehicle of mass $M$ is moving on a rough horizontal road with momentum $p$. If the coefficient of friction between the road and the tyres of the vehicle is $\mu_k$,what is the distance travelled by the vehicle before it comes to rest?

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