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Motion In Straight Line

Question
CBSEENPH11026277
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A stone is weighing 1 kg and sliding on ice with a velocity of 2 m/s is stopped by friction in 10 s. The force of friction (assuming it to be constant) will be

  • - 20 N

  • -0.2 N

  • 0.2 N

  • 20 N

Solution

B.

-0.2 N

u = 2 m/s

v = 0

f = 0s

The kinematic eqution is

    v = u - αt      

Where u is the initial velocity 

          v - final velocity

           α - acceleration

           t - time

⇒      α = v - ut

               = 0 - 210

         α = - 15

⇒      α = -0.2 m/s2

Friction force =  m α

                       = 1 × (-0.2)

                        = -0.2 N

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