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. A stone of 1 kg is thrown with a velocity of 20 m/s 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? * 1 point A) Zero B) 20N C) -4N D) 100N?
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. A stone of 1 kg is thrown with a velocity of 20 m/s across the froze...
Calculating the Force of Friction

Given:
- Mass of the stone (m) = 1 kg
- Initial velocity of the stone (u) = 20 m/s
- Distance travelled by the stone (s) = 50 m

Step 1: Calculate the acceleration of the stone
- Using the equation of motion: v^2 = u^2 + 2as
- Final velocity (v) = 0 m/s (stone comes to rest)
- Substituting the values: 0 = (20)^2 + 2a(50)
- Solving for acceleration (a): a = -4 m/s^2

Step 2: Calculate the force of friction
- The force of friction can be calculated using the equation: F = ma
- Substituting the values: F = 1 kg * (-4 m/s^2) = -4 N
Therefore, the force of friction between the stone and the ice is -4 N. Since the force of friction is in the opposite direction to the motion of the stone, it is negative, indicating that it acts in the direction opposite to the velocity of the stone. This negative value signifies that the force of friction is acting to decelerate the stone until it comes to a stop.
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. A stone of 1 kg is thrown with a velocity of 20 m/s across the froze...
Option c
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. A stone of 1 kg is thrown with a velocity of 20 m/s 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? * 1 point A) Zero B) 20N C) -4N D) 100N?
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