Given :-
- A 20kg block at an angle 53⁰ in an inclined plane is released from rest .
- [tex]\mu_s = 0.3 \ \& \ \mu_k = 0.2[/tex]
To Find :-
- Would the block move ?
- If it moves what is its speed after it has descended a distance of 5m down the plane .
Solution :-
For figure refer to attachment .
So the block will move if the angle of the inclined plane is greater than the angle of repose . We can find it as ,
[tex]\longrightarrow \theta_{repose}= tan^{-1}(\mu_s)[/tex]
Substitute ,
[tex] \longrightarrow \theta_{repose}= tan^{-1}( 0.2)[/tex]
Solve ,
[tex]\longrightarrow\underline{\underline{\theta_{repose}= 16.6^o }}[/tex]
Hence ,
[tex]\longrightarrow\theta_{plane}>\theta_{repose}[/tex]
Hence the block will slide down .
Now assuming that block is released from the reset , it's initial velocity will be 0m/s .
And the net force will be ,
[tex]\longrightarrow F_n = mgsin53^o - \mu_k N [/tex]
Substitute, N = mgcos53⁰ ( see attachment)
[tex] \longrightarrow ma_n = mgsin53^o - \mu_k mgcos53^o [/tex]
Take m as common,
[tex] \longrightarrow\cancel{m }(a_n) = \cancel{m}( gsin53^o - \mu gcos53^o)[/tex]
Simplify ,
[tex] \longrightarrow a_n = gsin53^o - \mu_k g cos53^o[/tex]
Substitute the values of sin , cos and g ,
[tex] \longrightarrow a_n = 10( 0.79 - 0.2 (0.6)) [/tex]
Simplify ,
[tex] \longrightarrow a_n = 10 ( 0.79 - 0.12 ) \\\\ \longrightarrow a_n = 10 (0.67)\\\\ \longrightarrow \underline{\underline{a_n = 6.7 m/s^2}}[/tex]
Now using the Third equation of motion namely,
[tex] \longrightarrow2as = v^2-u^2[/tex]
Substituting the respective values,
[tex] \longrightarrow2(6.7)(5) = v^2-(0)^2 [/tex]
Simplify and solve for v ,
[tex] \longrightarrow v^2 = 67 m/s\\\\\longrightarrow v =\sqrt{67} m/s \\\\\longrightarrow\underline{\underline{ v = 8.18 m/s }} [/tex]
Hence the velocity after covering 5m is 8.18 m/s .
