Question

Find an expression for the time taken by a body projected up along a rough inclined
plane of length { with an initial velocity u.

Answer 1

Answer:

When an object rests on a surface like the ramp, the ramp exerts a force called 'normal force' on the object, and this force is greater when the angle of incline is smaller. ... If the ramp is steep, the force of gravity can more easily overcome the force of friction

Explanation:

hope its help ;)

Answer 2

To derive:

Time taken by a body projected along a rough inclined plane with an initial velocity u.

Solution:

Let us consider that the elevation of the plane is $\theta$ and the coefficient of friction between the body and the plane is $\mu$.

So, frictional acceleration acting downward

$\therefore \: a_{f} = \mu(normal \: component \: of \: g)$

$= > \: a_{f} = \mu \{g \cos( \theta) \}$

So, According to FBD of the body:

$\therefore \: a_{net} = g \sin( \theta) + a_{f}$

$= > \: a_{net} = g \sin( \theta) + \mu g \cos( \theta)$

$= > \: a_{net} = g \bigg \{ \sin( \theta) + \mu \cos( \theta) \bigg \}$

Now , applying equations of kinematics:

$\therefore \: v = u - ( a_{net})t$

$= > \: 0= u - ( a_{net})t$

$= > \: 0= u - g \bigg \{ \sin( \theta) + \mu \cos( \theta) \bigg \}t$

$= > \: g \bigg \{ \sin( \theta) + \mu \cos( \theta) \bigg \}t = u$

$= > \: t = \dfrac{u}{g \bigg \{ \sin( \theta) + \mu \cos( \theta) \bigg \}}$

So, final answer is:

$\boxed{ \bold{\: t = \dfrac{u}{g \bigg \{ \sin( \theta) + \mu \cos( \theta) \bigg \}}}}$