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`        A block of weighing 130 is an incline , whose slope is 5 vertical to 12 horizontal. It`s initial velocity down the incline is 2.4 m/s. What will be the velocity 5 sec later? Take coefficient of friction at contact surface=0.3.`
3 years ago

```							If the mass of the body is taken in kg then,Given:-Height of incline=5m        :-Base of the incline=12m        :-w=0.3 (Coefficient of friction)        therefore SinA=5/13    (A=Base Angle)CosA=12/13Now, As per the free body diagram of the body,    1)mg downwards(gravity)so components of it, mgSinA Downwards along the incline                               and mgCosA downwards perpendicular the contact portion the component mgCosA gives rise to Normal Reaction (N) of same magnitude but opposite direction............ According to the formula of friction f=wNso, f= (0.3)x(130x10x12/13)  f=360 Newton The forward driving force (mgSinA) is (130x10x5/13)= 500 Newton  So, the resultant force on the body in (500-360)= 140 Newton down the incline......So, Force ma=140   Hence a= (140/m) = (140/130) =(14/13)  Applying Kinematical Equationv = u + atv = (2.4) + (14/13)(5)v = 7.79 (approx.)  Thanks And HOPE you understood it,ALL THE BEST........
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3 years ago
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