Question

The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vertical circle. What is the maximum speed the ball can have at the top of the circle? 3.74 m/s 9.67 m/s 5.86 m/s None of the above

Answer 1

v=5.86 m/s

Step-by-step explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

`F=(mv^2)/(r)`

v is the maximum speed

`v=\sqrt{(Fr)/(m)} \\\\v=\sqrt{(15* 0.8)/(0.35)} \\\\v=5.86\ m/s`

Hence, the maximum speed of the ball is 5.86 m/s.