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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]
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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