Question

$\huge \bf༆ question ༄$


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 ~

There's no picture in the question, you are free to make it yourself as per the given information .

Thanks for Answering ~ ​

Answer 1

Refer to the attachment for calculations

${\boxed{\Large{\mathfrak{s=\dfrac{v^2}{4gsin\Theta}}}}}$

How $\mu$gcos$\Theta$ turned to g$\sf sin\Theta$?

Ans:-

$\\ \sf\longmapsto \mu=tan\Theta$

Put in expression

$\\ \sf\longmapsto \mu g cos\Theta$

$\\ \sf\longmapsto tan\Theta g cos\Theta$

$\\ \sf\longmapsto \dfrac{sin\Theta}{cos\Theta}g cos\Theta$

$\\ \sf\longmapsto gsin\Theta$

Answer 2

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