<u>Given</u><u> </u><u>:</u><u>-</u>
- A 20kg block at an angle 53⁰ in an inclined plane is released from rest .
![\mu_s = 0.3 \ \& \ \mu_k = 0.2](https://tex.z-dn.net/?f=%5Cmu_s%20%3D%200.3%20%5C%20%5C%26%20%5C%20%5Cmu_k%20%3D%200.2)
<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 ,
![\longrightarrow\underline{\underline{\theta_{repose}= 16.6^o }}](https://tex.z-dn.net/?f=%5Clongrightarrow%5Cunderline%7B%5Cunderline%7B%5Ctheta_%7Brepose%7D%3D%2016.6%5Eo%20%7D%7D)
Hence ,
![\longrightarrow\theta_{plane}>\theta_{repose}](https://tex.z-dn.net/?f=%5Clongrightarrow%5Ctheta_%7Bplane%7D%3E%5Ctheta_%7Brepose%7D)
<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,
![\longrightarrow\cancel{m }(a_n) = \cancel{m}( gsin53^o - \mu gcos53^o)](https://tex.z-dn.net/?f=%20%5Clongrightarrow%5Ccancel%7Bm%20%7D%28a_n%29%20%3D%20%5Ccancel%7Bm%7D%28%20gsin53%5Eo%20-%20%5Cmu%20gcos53%5Eo%29)
Simplify ,
![\longrightarrow a_n = gsin53^o - \mu_k g cos53^o](https://tex.z-dn.net/?f=%20%5Clongrightarrow%20a_n%20%3D%20gsin53%5Eo%20-%20%5Cmu_k%20g%20cos53%5Eo)
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 ,
![\longrightarrow v^2 = 67 m/s\\\\\longrightarrow v =\sqrt{67} m/s \\\\\longrightarrow\underline{\underline{ v = 8.18 m/s }}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%20v%5E2%20%3D%2067%20m%2Fs%5C%5C%5C%5C%5Clongrightarrow%20v%20%3D%5Csqrt%7B67%7D%20m%2Fs%20%5C%5C%5C%5C%5Clongrightarrow%5Cunderline%7B%5Cunderline%7B%20v%20%3D%208.18%20m%2Fs%20%7D%7D%20)
<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>