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Manipal MET2010PhysicsMotion in One Dimension

A body is coming with a velocity of 72 ~km / h on a rough horizontal surface of coefficient of friction 0.5 . If the acceleration due to gravity is 10 ~m / s ^2 , find the minimum distance it can be stopped.

Options

  1. A400 m
  2. B40 m
  3. C0.40 m
  4. D4 m

Correct answer

B. 40 m

Step-by-step solution

aligned & Given, u=72 ~km / h =20 ~m / s & a= g=0.5 10 ~m / s ^2 & From, v^2=u^2-2 a s aligned (0)^2=(20)^2-2 0.5 10 s s= 20 20 2 0.5 10 or s=40 ~m

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