SELECT * FROM question_mgmt as q WHERE id=4138 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=8 AND subId=16 AND chapterId=189 and ex_no='3' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 9th science Force and Laws of Motion a stone of 1 kg is thrown with a velocity of 20 m

Question:

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?

Answer:

Mass of the stone = 1 kg

Initial velocity of the stone, u = 20 m/s

Final velocity of the stone, v = 0

Distance travelled by the stone, s = 50 m

Using the third equation of motion:

v2 = u2 + 2as

Where,

Acceleration, a

(0)2 = (20)2 + 2 × a × 50

a = −4 m/s2

Force, F = Mass × Acceleration

F = ma

F = 1 × (− 4) = −4 N

The negative sign shows that force of friction is in the opposite direction of motion , the force of friction between the stone and the ice is −4 N.


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