Question

9. A 0.25-kg mass is attached to a string and swung in a vertical circle whose radius is 0.75 m. At the bottom of the circle, the mass is observed to have a speed of 10 m/s. What is the magnitude of the tension in the string at that point

Answer 1

Answer:

T =  33.34 N

Explanation:

Given that,

Mass attached to the spring, m = 0.25 kg

Radius of the vertical circle, r = 0.75 m

Speed of the mass, v = 10 m/s

We need to find the magnitude of the tension in the string at that point. We know that the tension in the string is equal to the centripetal force acting on it. It can be given by

$T=\dfrac{mv^2}{r}\\\\T=\dfrac{0.25\times (10)^2}{0.75}\\\\T=33.34\ N$

So, the tension in the string is 33.34 N.