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 traveling a distance of 50 m. What is the force of friction between the stone and the ice?
Solution and Explanation
Mass of stone m = 1 kg, initial velocity u = 20 ms-1, Final velocity v= O (Therefore, the stone comes to rest), distance covered s= 50 m.
From third equation of motion,
v2 = u2 + 2as
(0)2 = (20)2 + 2a(50)
100a=-400 a=-4ms-2
Here negative sign shows that there is retardation in the motion of the stone. Force of friction between stone and ice = Force required to stop the stone = ma = 1 x -4 = -4N OR 4N
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Concepts Used:
Laws of Motion
The laws of motion, which are the keystone of classical mechanics, are three statements that defined the relationships between the forces acting on a body and its motion. They were first disclosed by English physicist and mathematician Isaac Newton.
Newton’s First Law of Motion
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s Second Law of Motion
Newton's 2nd law of motion deals with the relation between force and acceleration. According to the second law of motion, the acceleration of an object as built by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Newton’s Third Law of Motion
Newton's 3rd law of motion states when a body applies a force on another body that there is an equal and opposite reaction for every action.