<u>Given</u><u> </u><u>:</u><u>-</u>
- A 20kg block at an angle 53⁰ in an inclined plane is released from rest .

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
- Would the block move ?
- If it moves what is its speed after it has descended a distance of 5m down the plane .
<u>Solution</u><u> </u><u>:</u><u>-</u>
For figure refer to attachment .
So the block will move if the angle of the inclined plane is greater than the <u>angle</u><u> of</u><u> </u><u>repose</u><u> </u>. We can find it as ,
Substitute ,
Solve ,

Hence ,

<u>Hence</u><u> the</u><u> </u><u>block</u><u> will</u><u> slide</u><u> down</u><u> </u><u>.</u>
Now assuming that block is released from the reset , it's <u>initial</u><u> </u><u>velocity </u> will be 0m/s .
And the net force will be ,
Substitute, N = mgcos53⁰ ( see attachment)
Take m as common,

Simplify ,

Substitute the values of sin , cos and g ,
Simplify ,
Now using the <u>Third </u><u>equation</u><u> </u><u>of</u><u> motion</u><u> </u>namely,
Substituting the respective values,
Simplify and solve for v ,

<u>Hence</u><u> the</u><u> </u><u>velocity</u><u> after</u><u> </u><u>covering</u><u> </u><u>5</u><u>m</u><u> </u><u>is </u><u>8</u><u>.</u><u>1</u><u>8</u><u> </u><u>m/</u><u>s </u><u>.</u>