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Grade: 12th pass
        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.
2 years ago

Answers : (1)

Suyash Verma
12 Points
							
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=12/13
Now, 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=wN
so, 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 Equation
v = u + at
v = (2.4) + (14/13)(5)
v = 7.79 (approx.)
 
 
Thanks And HOPE you understood it,
ALL THE BEST........
2 years ago
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