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[tex]\huge \bf༆ question ༄[/tex]


A block slides down an inclined plane of slope angle Φ with constant velocity. if it is then projected up the same plane with an initial speed " v " . the distance in which it will come to rest is ?

Find distance ~

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Thanks for Answering ~ ​

Respuesta :

Refer to the attachment for calculations

[tex]{\boxed{\Large{\mathfrak{s=\dfrac{v^2}{4gsin\Theta}}}}}[/tex]

How [tex]\mu[/tex]gcos[tex]\Theta[/tex] turned to g[tex]\sf sin\Theta[/tex]?

Ans:-

[tex]\\ \sf\longmapsto \mu=tan\Theta[/tex]

Put in expression

[tex]\\ \sf\longmapsto \mu g cos\Theta[/tex]

[tex]\\ \sf\longmapsto tan\Theta g cos\Theta[/tex]

[tex]\\ \sf\longmapsto \dfrac{sin\Theta}{cos\Theta}g cos\Theta[/tex]

[tex]\\ \sf\longmapsto gsin\Theta[/tex]

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cutiepiesri

As the block is sliding down with constant velocity. Therefore friction force along the slope must be equal to the component of gravity along the slope.

μN = mgsin0

umgcose mgsine

=

μ = tane

When thrown up: Initial Velocity is Vo

Acceleration (both friction and

component of gravity acting downwards the slope as relative motion

is upwards) is µN + mgsin0 = tanmgcose +

mgsin0 = 2mgsin

Final Velocity is zero.

Therefore distance traveled up the incline is:

v² - u² = 2as

⇒ 0-V₂² =2(-2mgsin0)s

S = 4mgsin

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