Question

question 18
a certain string just breaks when it is under 25 n of tension. a boy uses this string to whirl a 0.75 - kg stone in a horizontal circle of radius 2.0 m. the boy continuously increases the speed of the stone. at approximately what speed will the string break?
15 m/s
8.2 m/s
6.4 m/s
18 m/s
12 m/s

Explanation:

Step1: Identify centripetal - force formula

The centripetal force $F = \frac{mv^{2}}{r}$, where $F$ is the centripetal force, $m$ is the mass of the object, $v$ is the speed of the object, and $r$ is the radius of the circular path. The tension in the string provides the centripetal force, and when the string breaks, $F = T_{max}=25\ N$, $m = 0.75\ kg$, and $r = 2.0\ m$.

Step2: Rearrange formula for speed

From $F=\frac{mv^{2}}{r}$, we can solve for $v$. First, multiply both sides by $r$: $Fr=mv^{2}$. Then divide both sides by $m$: $v^{2}=\frac{Fr}{m}$. Take the square - root of both sides: $v=\sqrt{\frac{Fr}{m}}$.

Step3: Substitute values

Substitute $F = 25\ N$, $r = 2.0\ m$, and $m = 0.75\ kg$ into the formula $v=\sqrt{\frac{Fr}{m}}$. So $v=\sqrt{\frac{25\times2}{0.75}}=\sqrt{\frac{50}{0.75}}=\sqrt{\frac{50}{\frac{3}{4}}}=\sqrt{\frac{200}{3}}\approx8.2\ m/s$.

Answer:

$8.2\ m/s$