- 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?
Answers
Answer :
›»› Friction force friction between the stone and the ice is -4 N.
Given :
- Mass of a stone = 1 kg
- Initial velocity of a stone = 20 m/s
- Final velocity of a stone = 0 m/s
- Distance travelled by a stone = 50 m
To Find :
- Friction force friction between the stone and the ice = ?
Required Solution :
⪼ Final velocity (v) = 0 m/a (because stone come to rest)
Here in this question, we have to find Friction force friction between the stone and the ice. So, firstly we need to find Acceleration of a stone, after that we will find Friction force friction between the stone and the ice on the basis of conditions given above.
From third equation of motion
→ v² = u² + 2as
→ 0² = 20² + 2 × a × 50
→ 0 = 400 + 2a × 50
→ 0 = 400 + 100a
→ 0 - 400 = 100a
→ -400 = 100a
→ a = -400/100
→ a = -4 m/s²
【NOTE : Negative sign of indicates retardation/deceleration.】
Now, we have two elements that used in formula, Mass and Acceleration of a stone.
- Mass of a stone (m) = 1 kg
- Acceleration of a stone (a) = -4 m/s²
And we need to find Friction force friction between the stone and the ice.
From second law of Newton
→ F = ma
→ F = 1 × (-4)
→ F = -4 N
【NOTE : Negative sign of force indicates retarding force.】
║Hence, the Friction force friction between the stone and the ice is -4 N.║