(D) Given: Mass $m = 5\, kg$,Applied force $F = 24\, N$,Coefficient of kinetic friction $\mu_k = 0.4$,Acceleration due to gravity $g = 9.8\, m/s^2$.Fi

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(D) Given: Mass $m = 5\, kg$,Applied force $F = 24\, N$,Coefficient of kinetic friction $\mu_k = 0.4$,Acceleration due to gravity $g = 9.8\, m/s^2$.First,calculate the kinetic frictional force: $f_k = \mu_k \cdot m \cdot g = 0.4 \times 5 \times 9.8 = 19.6\, N$.Since the applied force $F = 24\, N$ is greater than the frictional force $f_k = 19.6\, N$,the block will accelerate.The net force acting on the block is $F_{net} = F - f_k = 24 - 19.6 = 4.4\, N$.Using Newton's second law,$F_{net} = ma$,the acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{4.4}{5} = 0.88\, m/s^2$.

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

Medium

$A$ block of mass $5\, kg$ is on a rough horizontal surface and is at rest. Now a force of $24\, N$ is imparted to it with negligible impulse. If the coefficient of kinetic friction is $0.4$ and $g = 9.8\, m/s^2$,then the acceleration of the block is ........ $m/s^2$.

A
$0.26$
B
$0.39$
C
$0.69$
D
$0.88$

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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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MediumAIPMT 1992

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$A$ block of mass $m$ is moving on a rough horizontal surface. The coefficient of kinetic friction between the block and the surface is $\mu_{k}$. The net force exerted by the surface on the block is (where $g$ is the acceleration due to gravity).

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The magnitude of acceleration of a block moving with a speed of $10\,m/s$ on a rough surface is $.........\,m/s^2$,if the coefficient of friction is $0.2$.

Easy

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$A$ block of mass $5 \,kg$ is pulled by a force $F$ as shown in the figure. If the coefficient of friction is $0.1$, then the force needed to accelerate the block to $3 \,m/s^2$ to the right is close to (in $\,N$)

EasyTS EAMCET 2018

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$A$ block of mass $5\, kg$ is on a rough horizontal surface and is at rest. Now a force of $24\, N$ is imparted to it with negligible impulse. If the coefficient of kinetic friction is $0.4$ and $g = 9.8\, m/s^2$,then the acceleration of the block is ........ $m/s^2$.

Similar Questions

$A$ conveyor belt is moving at a constant speed of $2\, m s^{-1}$. $A$ box is gently dropped on it. The coefficient of friction between them is $\mu = 0.5$. The distance that the box will move relative to the belt before coming to rest on it,taking $g = 10\, m s^{-2}$,is ........... $m$.

Consider a car moving along a straight horizontal road with a speed of $72 \, km/h$. If the coefficient of kinetic friction between the tyres and the road is $0.5$,the shortest distance in which the car can be stopped is ........ $m$. $[g = 10 \, m/s^2]$

$A$ block of mass $m$ is moving on a rough horizontal surface. The coefficient of kinetic friction between the block and the surface is $\mu_{k}$. The net force exerted by the surface on the block is (where $g$ is the acceleration due to gravity).

The magnitude of acceleration of a block moving with a speed of $10\,m/s$ on a rough surface is $.........\,m/s^2$,if the coefficient of friction is $0.2$.

$A$ block of mass $5 \,kg$ is pulled by a force $F$ as shown in the figure. If the coefficient of friction is $0.1$, then the force needed to accelerate the block to $3 \,m/s^2$ to the right is close to (in $\,N$)

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