A block of mas 0.5 kg has an initial velocity of $10 m s^{- 1}$ down an inclined rough plane of angle 30, the coefficient of friction between the block and inclined surface being 02. the velocity of the block after it travels a distance of 10 m is.

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Solution

From above image
$F_{n e t} = m g s i n 30^{\circ} - μ m g c o s 30^{\circ}$
so, $a_{n e t} = 5 - \sqrt 3$

From A to B
u = 10m/s v=? a=5-√3 S = 10m
$v^{2} = u^{2} + 2 a S$
$v^{2} = 100 + 20 (5 - \sqrt 3)$
$v^{2} = 100 + 100 - 20 \sqrt 3$
$v = \sqrt{200 - 20 \sqrt 3}$

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A block of mas 0.5 kg has an initial velocity of 10ms−1 down an inclined rough plane of angle 30, the coefficient of friction between the block and inclined surface being 02. the velocity of the block after it travels a distance of 10 m is.

From above image Fnet=mgsin30∘−μmgcos30∘so, anet=5−√3From A to Bu = 10m/s v=? a=5-√3 S = 10mv2=u2+2aSv2=100+20(5−√3)v2=100+100−20√3v=√200−20√3

A block of mas 0.5 kg has an initial velocity of $10 m s^{- 1}$ down an inclined rough plane of angle 30, the coefficient of friction between the block and inclined surface being 02. the velocity of the block after it travels a distance of 10 m is.

From above image
$F_{n e t} = m g s i n 30^{\circ} - μ m g c o s 30^{\circ}$
so, $a_{n e t} = 5 - \sqrt 3$

From A to B
u = 10m/s v=? a=5-√3 S = 10m
$v^{2} = u^{2} + 2 a S$
$v^{2} = 100 + 20 (5 - \sqrt 3)$
$v^{2} = 100 + 100 - 20 \sqrt 3$
$v = \sqrt{200 - 20 \sqrt 3}$