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In the given figure if the block is projected with 20 m/s in upward direction then find the distance cover (in meters) to come at rest (Approx) (g=10m/s2)

Option: 1

26


Option: 2

15


Option: 3

12


Option: 4

20


Answers (1)

best_answer

As we learnt

Sopping of Block on Inclined Road -

a=g[sin\theta+\mu cos\theta]

V^{2}=u^{2}-2aS

0=u^{2}-2g[sin\theta+\mu cos \theta]S

S=\frac{u^{2}}{2g(sin\theta+\mu cos\theta)}

- wherein

S= distance traveled

\mu= coefficient of friction

V = Final velocity

u = Initial velocity

 

 When we project a block on an inclined plane then retardation of block 

 

a_{net} =gsin\theta+\mu g\cos\theta

a = 10 *\frac{3}{5}+0.2*10*\frac{4}{5}

a= 7.6 m/s^{2}

v^{2}= u ^{2}-2as \Rightarrow 0= u ^{2}-2as

s=\frac{u ^{2}}{2\times a}=\frac{20*20}{2*7.6}=26 m

 

 

 

Posted by

Riya

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