Question

A string under a tension of 68 N is used to whirl a rock in a horizontal circle of radius 3.7 m at a speed of 16.53 m/s. The string is pulled in, and the speed of the rock increases. When the string is 0.896 m long and the speed of the rock is 71.5 m/s, the string breaks. What is the breaking strength of the string? Answer in units of N.

Answer 1

Answer:

$F = 5253.7 N$

Explanation:

As we know that tension force in the string will be equal to the centripetal force on the string

so we will have

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

now we have

$68 = \frac{m(16.53^2)}{3.7}$

now we have

$68 = 73.8 m$

$m = 0.92 kg$

now when string length is 0.896 m and its speed is 71.5 m/s then we will have

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

$F = \frac{0.92(71.5^2)}{0.896}$

$F = 5253.7 N$

Answer 2

Answer:5249.18 N

Explanation:

Given

When Tension T is 68 N velocity v=16.53 m/s

radius r=3.7 m

as the rock is in the horizontal position thus tension will provide the centripetal acceleration

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

$68=\frac{m\times 16.53^2}{3.7}$

m=0.92 kg

when Velocity is 71.5 m/s and r=0.896 m string breaks

therefore breaking strength of string is

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

$T=\frac{0.92\times 71.5^2}{0.896}$

$T=5249.18 N$