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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 ~ ​

Sagot :

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]

View image Аноним
View image Аноним
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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