Answer to Question #92767 in Mechanics | Relativity for Sridhar

Question

A rough inclined plane BCE of height (256)m is kept on a rectangular wooden block ABCD of height 10 m. as shown in the figure. A small block is allowed to slide down from the top E of the inclined plane. The coefficient of kinetic friction between the block and the inclined plane is 18 and the angle of inclination of the inclined plane is sin−1(0.6). If the small block finally reaches the ground at point F. then DF = ?
(Acceleration due to gravity = 10ms−2
Ans20/3

Answer 1

According to the Newton's second law, we can write

$mg × \sinθ – f = ma (1)$

where f is the friction force

The friction force is given by formula

$f=µ mg × \cosθ (2)$

We put (2) in (1) and get formula for acceleration

$a= g (\sinθ – µ × \cosθ) (3)$

In our case, we have

$\sinθ=0.6$

$\cosθ=0.8$

µ=0.125

Using (3) we get

a=0.5 m/s²

The movement on EC is accelerated and we can write

$2as=v^2 (4)$

where v is the speed at a point C

Using (4) we get

$v=\sqrt {2as} (5)$

where s=EC, a=0.5 m/s²

From figure we have

$EC=\sqrt{{EB}^2 + {BC}^2 } (6)$

where

$EB=\frac {25}{6}$

BC=10

The movement on CF is a body movement thrown horizontally

In this sense, we can write

$DE=v×\sqrt{\frac {2×CD}{g}} (7)$

where CD=10

Using (5), (6), (7) we get

$DE=\frac {20}{3}$

Answer

$\frac {20}{3}$