Question

a block of mass 0.5kg is placed on plane making angle 30 degree with horizontal.If the coefficient of friction between the surface of contact of the body and the plane is 0.2,what force is required to keep the body sliding down with uniform velocity?
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Answer 1

Answer:

θ = 30°

mass of the block (m) = 0.5kg

coefficient of friction (μ) = 0.2

force need to move the block (F) = ?

here,

F = mg. sin θ - f [ Here f is the

opposing force,

friction ]

=> F = mg. sin θ - μR [f = μR]

=> F = mg. sin θ - μ.mg. cos θ

[R=mg.cos θ]

=> F = mg(sin θ - μ. cos θ )

=> F = 0.5 X 10 (sin 30 - 0.2 X cos 30)

=> F = 5(1/2 - 0.2 X √3/2)

=> F = 5[(5-√3)/10] {√3 = 1.7}

=> F = 5 X 0.33

=> F = 1.65 N

Hence, 1.65 N force is required to slide down the block.

Answer 2

$\huge{\mathfrak{\underline{Answer:-}}}$

$\large{\sf{Here \: {\Theta} \: = 30 ^{\circ}}}$

$\large{\sf{ Mass \: of \: the \: block \: (m) \: = \: 0.5kg}}$

$\large{\sf{Cofficient \: of \: friction \: ({\mu}) \: = \: 0.2}}$

$\huge{\boxed{\sf{F \: = \: mg.sin{\Theta} \: - \: f }}}$

$\large{\sf{F \: = \: mg \: </p><p>Sin{\Theta} \: - {\mu}R}}$

$\large{\sf{F \: = \: mg \: Sin{\Theta} \: - \: {\mu}mg Cos{\Theta}}}$

$\large{\sf{ F \: = \: mg (sin{\Theta} - {\mu} Cos {\Theta}}})$

$\large{\sf{F \: = \: 0.5 {\times}{10}(Sin30^{\circ} - 0.2 Cos 30^{\circ}}})$

$\large{\sf{F \: = \: 5(1/2 - 0.2 {\times}{\sqrt{3}}/2}})$

$\large{\sf{F \: = \: 5(5 - {\sqrt{3}}/10 }})$

$\large{\sf{ F \: = \: 0.33 {\times}{5}}}$

$\large{\sf{F \: = \: 1.65 N}}$

$\large{\boxed{\sf{F \: = \: 1.65 N}}}$