A block of mass 2 , kg is placed on a surface | vedclass

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Question

(A) The maximum static frictional force (limiting friction) is given by $f_{max} = \mu N = \mu mg$. Given: $m = 2 \, kg$,$\mu = 0.4$,and $g = 9.8 \, m

Vedclass · Accepted Answer

$2.5$

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(A) The maximum static frictional force (limiting friction) is given by $f_{max} = \mu N = \mu mg$. Given: $m = 2 \, kg$,$\mu = 0.4$,and $g = 9.8 \, m/s^2$. $f_{max} = 0.4 	imes 2 	imes 9.8 = 7.84 \, N$. Since the applied force $F = 2.5 \, N$ is less than the limiting friction $f_{max} = 7.84 \, N$,the block will not move. According to the law of static friction,the frictional force adjusts itself to be equal to the applied force when the body is at rest. Therefore,the frictional force $f = F = 2.5 \, N$.

$A$ block of mass $2 \, kg$ is placed on a surface with a coefficient of friction $\mu = 0.4$. If a force of $2.5 \, N$ is applied to it,the frictional force will be ........ $N$.

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