Question

there is an inclined plane with length 13m height 5m and coefficient of friction 1/3 with what initial velocity a body is to be slided up the plane so that it just comes to rest at the top of the inclined plane​

Answer 1

Given:

There is an inclined plane with length 13m height 5m and coefficient of friction ⅓.

To find:

What initial velocity a body is to be slided up the plane so that it just comes to rest at the top of the inclined plane ?

Calculation:

Let the angle of inclined plane be $\theta$.

Now, let initial velocity be u , and final velocity at top of inclined plane will be zero :

Now , the net deceleration experienced by object on inclined plane will be :

$\therefore \: a = - g \bigg \{ \sin( \theta) + \mu \cos( \theta) \bigg \}$

Applying 3rd equation of kinematics:

$\therefore \: {v}^{2} = {u}^{2} + 2as$

$\implies \: {(0)}^{2} = {u}^{2} + 2 as$

$\implies \: {u}^{2} = - 2 as$

$\implies \: {u}^{2} = - 2 [ - g\{ \sin( \theta) + \mu \cos( \theta) \}]s$

$\implies \: {u}^{2} = 2 g\{ \sin( \theta) + \mu \cos( \theta) \}s$

$\implies \: {u}^{2} = 2 g\{ \dfrac{5}{13} + \mu ( \dfrac{12}{13} ) \}13$

$\implies \: {u}^{2} = 2 g\{ 5 + \mu (12 ) \}$

$\implies \: {u}^{2} = 2 g\{ 5 + (12 \times \dfrac{1}{3} ) \}$

$\implies \: {u}^{2} = 2 g\{ 5 + 4\}$

$\implies \: {u}^{2} = 18g$

$\implies \: {u}^{2} = 18 \times 10$

$\implies \: {u}^{2} = 180$

$\implies \: u = \sqrt{180}$

$\implies \: u = 13.41 \: m {s}^{ - 1}$

So, the required velocity is 13.41 m/s.