Question

A body of mass 2 kg moves in a horizontal circular path of radius 5m. In an instant its speed is \(2\sqrt5\) m/s and is increasing at the rate of 3 m/s².The magnitude of the force acting on the body at the instant is,

• A. 6N
• B. 8N
• C. 14N
• D. 10N

Answer 1

The Correct Option is D

The magnitude of the force acting on the body at that instant is 10 N due to the combination of the centripetal force required to keep the body moving in a circular path (directed towards the center of the circle) and the tangential force responsible for increasing the speed of the body.

The correct answer is option (D): 10N

Answer 2

To determine the magnitude of the force acting on the body moving in a horizontal circular path, we need to consider both the centripetal force (due to circular motion) and the tangential force (due to the change in speed). Here are the steps:
1. Centripetal Force:
  - The centripetal force \(F_c\) is given by:
    \[   F_c = \frac{mv^2}{r}  \]
  - Where \(m\) is the mass of the body, \(v\) is the speed, and \(r\) is the radius of the circular path.
  - Substituting the given values:
    \[   m = 2\, \text{kg}, \quad v = 2\sqrt{5}\, \text{m/s}, \quad r = 5\, \text{m} \]
  - Calculate \(v^2\):
    \[  v^2 = (2\sqrt{5})^2 = 4 \times 5 = 20\, \text{m}^2/\text{s}^2\]
  - Now, calculate the centripetal force:
    \[   F_c = \frac{2 \times 20}{5} = \frac{40}{5} = 8\, \text{N}  \]
2. Tangential Force:
  - The tangential force \(F_t\) is due to the tangential acceleration \(a_t\):
    \[    F_t = ma_t  \]
  - Given \(a_t = 3\, \text{m/s}^2\), so:
    \[  F_t = 2 \times 3 = 6\, \text{N}  \]
3. Total Force:
  - The total force \(F\) is the vector sum of the centripetal force and the tangential force. Since these forces are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the total force:
    \[  F = \sqrt{F_c^2 + F_t^2}   \]
  - Substitute the values:
    \[  F = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\, \text{N}    \]
Conclusion
The magnitude of the force acting on the body at the given instant is \(10\, \text{N}\) So the correct Answer is Option (D):10N.