A stone of 1kg is thrown with a velocity of 20 ms^-1 across the frozen surface of a lake and comes to rest after traveling a distance of 50m. what is the force of friction between the stone and the ice?
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No wrong answer please
Answers
AnswEr :
- The force of friction between the stone and the ice is -4 N.
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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
Unknown :
- The force of friction between the stone and the ice?
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Calculations :
Here in this question, we are asked to find the force of friction between the stone and the ice, so firstly we need to find out the acceleration of the stone, after that we can easily find the force of friction between the stone and the ice on the basis of condition given above. Let's do it...!
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We can find the acceleration of the stone by using the third equation of motion, which says:
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- v² = u² + 2as
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[Where, v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance travelled by the stone.]
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So let's calculate acceleration (a) !
From third equation of motion
→ v² = u² + 2as
→ 0² = (20)² + 2 × a × 50
→ 0 = (20)² + 2 × a × 50
→ 0 = 400 + 2 × a × 50
→ 0 = 400 + 2a × 50
→ 0 = 400 + 100a
→ 0 - 400 = 100a
→ -400 = 100a
→ a = -400/100
→ a = -4
【NOTE : Negative sign of acceleration indicates retardation/deceleration.】
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Now, we have two elements that used in formula, Mass and Acceleration of a stone.
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- Mass of a stone (m) = 1 kg
- Acceleration of a stone (a) = -4 m/s²
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And we need to find the force of friction between the stone and the ice.
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Now, we can easily find the force of friction between the stone and the ice by using the second law of Newtown, which says:
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- F = ma
[Where, F is the force, m is the mass of the stone and a is the acceleration produced by the stone.]
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Let's find force (F) !
→ F = ma
→ F = 1 × (-4)
→ F = -4 N
【NOTE : Negative sign of force indicates retarding force.】
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Hence, the force of friction between the stone and the ice is -4N.
Solution -
We have,
- Initial velocity (u) = 20 ms-¹
- Final velocity (v) = 0
- Distance travelled (s) = 50 m
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Firstly, we will find the acceleration using the third equation of motion.
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⠀⠀⠀⠀⠀⠀★ v² = u² + 2as
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➝ (0)² = (20)² + 2 × a × 50
➝ 0 = 400 + 2a × 50
➝ 0 = 400 + 100a
➝ 100a = - 400
➝ a = -4 ms-²
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Now, we have to find the force of friction between the stone and the ice. For this we use the second law of Newton
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⠀⠀⠀⠀⠀⠀⠀★ F = m × a
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➝ Force = 1 × (-4)
➝ Force = -4N
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Thus, the force of friction between the stone and the ice is 4 Newtons. The negative sign shows that this force opposes the motion of stone.