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If a stone is tied to a string and whirled in a horizontal circle of radius 2 m

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Question: If a stone is tied to a string and whirled in a horizontal circle of radius 2 m at a speed of 4 m/s, what is the tension in the string if the mass of the stone is 1 kg?

Options:

  1. 2 N
  2. 4 N
  3. 8 N
  4. 10 N

Correct Answer: 8 N

Solution: 

Centripetal force = m(v²/r) = 1(4²/2) = 8 N. Tension in the string = 8 N.

If a stone is tied to a string and whirled in a horizontal circle of radius 2 m

Practice Questions

Q1
If a stone is tied to a string and whirled in a horizontal circle of radius 2 m at a speed of 4 m/s, what is the tension in the string if the mass of the stone is 1 kg?
  1. 2 N
  2. 4 N
  3. 8 N
  4. 10 N

Questions & Step-by-Step Solutions

If a stone is tied to a string and whirled in a horizontal circle of radius 2 m at a speed of 4 m/s, what is the tension in the string if the mass of the stone is 1 kg?
  • Step 1: Identify the given values. The mass of the stone (m) is 1 kg, the speed (v) is 4 m/s, and the radius (r) is 2 m.
  • Step 2: Use the formula for centripetal force, which is F = m(v²/r). This formula helps us find the force needed to keep the stone moving in a circle.
  • Step 3: Calculate v² (the speed squared). Here, v = 4 m/s, so v² = 4 * 4 = 16 m²/s².
  • Step 4: Divide v² by r (the radius). We have v² = 16 m²/s² and r = 2 m, so 16 / 2 = 8 m/s².
  • Step 5: Multiply the result by the mass (m). We have m = 1 kg, so the centripetal force is 1 kg * 8 m/s² = 8 N.
  • Step 6: The tension in the string is equal to the centripetal force needed to keep the stone moving in the circle, which is 8 N.
  • Centripetal Force – The force required to keep an object moving in a circular path, directed towards the center of the circle.
  • Tension in a String – The force exerted by a string when it is pulled tight by forces acting from opposite ends.
  • Newton's Second Law – The relationship between the force acting on an object, its mass, and its acceleration (F = ma).
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