Question

A stone with a mass of 0.70 kg is attached to one end of a string 0.70 m long. The string will break if its tension exceeds 65.0 N. The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed.

Answer 1

Answer:v=8.06 m/s

Explanation:

Given

mass of stone $m=0.7 kg$

length of string $L=0.7 m$

Maximum Tension $T=65 N$

the string will not break if Centripetal Force is less than maximum tension

$T=\frac{mv^2}{r}$

$65=\frac{0.7\times v^2}{0.7}$

$v^2=65$

$v=\sqrt{65}$

$v=8.06 m/s$