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Q 6.      A stone of 1 kg is thrown with a velocity of 20 m s-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

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By the equation of motion, we know that,

                                                              v^2\ =\ u^2\ +\ 2as

or                                                           0^2\ =\ 20^2\ +\ 2a\times 50                                           (Since ball comes to rest thus final velocity is zero.)

or                                                           a\ =\ \frac{-400}{100}\ =\ -4\ m/s^2 

The acceleration is negative. This implies that the force is opposing the motion.

         The force is given by :

                                                               F\ =\ ma

or                                                            F\ =\ 1\times \ (-4)\ =\ -4\ N

Hence the frictional force between stone and ice is 4 N. 

Posted by

Devendra Khairwa

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