Question 1188713
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A body {{{cross(slide)}}} <U>SLIDES</U> down {{{cross(an)}}} <U>AT</U> incline plane which {{{cross(make)}}} <U>MAKES</U> angle of 30° with the horizontal. 
Neglecting friction, calculate
a) The velocity of the body after sliding 10m from rest.
b) The time taken to slide 10m distance.
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<pre>
(a)  Notice that the height "h" is half of the hypotenuse, since the angle with horizont is 30.

     So, the height is  {{{10/2}}} = 5 meters.

     The change of the potential energy between the levels h = 5 meters and h = 0 if  P = mgh,

     where m is the mass of the body in kilograms and h = 10m/s^2 is the gravity acceleration.


     This change of the potential energy is spent to increase the kinematic energy of the body 

     from 0 to  {{{(mv^2)/2}}}.  where v is the magnitude of the velocity at the end.


     So,   {{{(mv^2)/2}}} = mgh.   It is nothing elese as the energy conservation law in this problem.


     Canceling "m" in both sides, you get  {{{v^2}}} = 2gh,  or   v = {{{sqrt(2gh)}}}.


     The last step is to substitute h = 5 m,  g = 10 m/s^2  into this formula, and you get

         v = {{{sqrt(2*10*5)}}} = {{{sqrt(100)}}} = 10 m/s.


<U>ANSWER</U>.  The final speed (the magnitude of the velocity) is 10 m/s.




(b)  In this problem, the speed rise from 0 to 10 m/s uniformly with time.

     
     Hence, the average speed is the average between 0 and 10 m/s,  which is 5 m/s.


     It implies that the time taken to slide is the full distance of 10 meters divided by the average speed of 5 m/s


        time taken to slide = {{{10/5}}} = 2 seconds.
</pre>

Solved.  &nbsp;&nbsp;//  &nbsp;&nbsp;Both questions are answered.