Grade 12th passMechanics

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.

anil chinni

9 Years agoGrade 12th pass

1 Answer

Suyash Verma

9 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=12/13

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........

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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.

anil chinni

1 Answer

Suyash Verma

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