Question

A block of mass 10kg is pushed by a force F on a horizontal rough plane is moving with acceleration 5 ms-2 When force is doubled, its acceleration becomes 18ms-2. Find the coefficient of friction between the block and rough horizontal plane. (g = 10m/s2).​

Answer 1

We know:

ma=F−μmg

Case 1:

5×10=F−100μ

Case 2:

18×10=2F−100μ

Solving simultaneously we get:

F=130N, μ=0.8

Answer 2

AnSwer :

On a rough horizontal plane accerlation of a block of mass'm'is given by

${ : \implies {\sf{ a = \dfrac{F}{m} - \mu _{k}g \: \: \: \: \: .......(1)}}}$

Initially, a = 5ms¯²

${: \implies{ \sf {5 = \dfrac{F}{10} - \mu _{k}(10)\: \: \: \: \: .......(2)}}}$

When force is doubled, a = 10ms¯²

${: \implies{ \sf {18= \dfrac{2F}{10} - \mu _{k}(10)\: \: \: \: \: .......(3)}}}$

now, let's multiple eq (2) with 2 .i.e.,

${: \implies{ \sf {2(5)= 2 \bigg(\dfrac{F}{10} - \mu _{k}(10) \bigg)}}}$

${: \implies{ \sf {10= \dfrac{F}{5} - 20 \mu _{k} \: \: \: \: ......(4)}}}$

now, let's subtract eq (4) with (3)

${: \implies{ \sf {18 - 10= \dfrac{2F}{10} - \dfrac{F}{5} - 10\mu_{k} - 20\mu _{k}}}}$

${: \implies{ \sf {8= \dfrac{2F - 2F}{10} + 10 \mu _{k} }}}$

${ : \implies {\sf {8 = 10 \mu_{k}}}}$

${ : \implies {\sf {\mu_{k} = \dfrac{8}{10} }}}$

${ : \implies {\sf {\mu_{k} = 0.8}}}$

thus, the coefficient of friction between the block and rough horizontal plane is 0.8

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