Question

on smooth inclind plane of inclination = 45 drgree a bodyb takes time 2s to reach the bottom .when plane is made rough (coffecient of friction =0.75) ,the same body will take time ?????????????????????????????????/

Answer 1

1. On smooth incline plane of inclination $= 45$ degree a body takes time 2s to reach the bottom. When plane is made rough (coefficient of friction=0.75), the same body will take time?

Let write down the equation of the motion of the body (Newton second law) and also write in projectives onto the $X$ - and $Y$ -axes.

$$\overrightarrow {m a} = \overrightarrow {m g} + \overrightarrow {N} + \overrightarrow {F}.$$

$\left\{ \begin{array}{l}ma = mg\sin \alpha -F\\ m\cdot 0 = -mg\cos \alpha +N \end{array} \right.$ where the friction force is $F = \mu N$

So, the acceleration of the body is constant:

$$a = g \sin \alpha - \frac {F}{m} = g \sin \alpha - \frac {\mu \cdot m g \cos \alpha}{m} = g (\sin \alpha - \mu \cos \alpha).$$

Let the length of the plane be $l$ .

We can find the time of the motion of the body with the constant acceleration:

$$t = \sqrt {\frac {2 l}{a}} = \sqrt {\frac {2 l}{g (\sin \alpha - \mu \cos \alpha)}}.$$

So, time of the motion in the case of smooth plane $(\mu = 0)$ and rough one

$$t = \sqrt {\frac {2 l}{g \sin \alpha}}, \quad t _ {1} = \sqrt {\frac {2 l}{2 g (\sin \alpha - \mu \cos \alpha)}}.$$

So, $\begin{array}{r}t_{1}\\ t \end{array} = \sqrt{\frac{2l}{2g(\sin\alpha - \mu\cos\alpha)}}:\sqrt{\frac{2l}{g\sin\alpha}}.$

As one can find out,

$$\boxed {t _ {1} = \frac {t}{\sqrt {1 - \mu} c t g \alpha}.}$$

Let check the dimension.

$\left[t_1\right] = s$

Let evaluate the quantity.

$$t _ {1} = \frac {2}{\sqrt {1 - 0 . 7 5} \cdot c t g 4 5 ^ {0}} = \frac {2}{\sqrt {1 - 0 . 7 5} \cdot 1} = 4 (s).$$

Answer: 4 s.