A stone with mass 0.80 kg is attached to one end of a string 0.90 m long. The string will break if its tension exceeds 60.0 N. The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. (a) Draw a free-body diagram of the stone. (b) Find the maximum speed the stone can attain without the string breaking.

(b) The stone attached with the string experiences both centripetal (towards the center) and centrifugal (away from the center) forces. The tension of the string counters the centrifugal force until it breaks.

We know that,

Centrifugal force = [tex]\frac{mv^2}{r}[/tex]

where,

[tex]m[/tex] = mass of the stone

[tex]v[/tex] = velocity of the stone

[tex]r[/tex] = length of the string

To find the maximum speed attained by the stone without the string breaking, we must equate:

[tex]\frac{mv^2}{r} =60[/tex]

or, [tex]v=\sqrt \frac{{r\times 60}}{m} } =\sqrt{\frac{0.90\times60}{0.80} } =8.2158m/s[/tex]

lhabdulsamirahmed lhabdulsamirahmed · Answer 2 · 2022-03-03T12:44:41+01:00

The maximum speed the stone can attain is 8.215 m/s

Data;

Mass = 0.80kg
radius of tangential path = 0.90m
Tension = 60N

Centripetal Force

The centripetal force of the stone can be calculated as

[tex]F_c = \frac{mv^2}{r} \\[/tex]

The tension acting on the string is given as

[tex]T = \frac{mv^2}{r} \\v^2 = \frac{Tr}{m} \\v = \sqrt{\frac{Tr}{m} }\\ v = \sqrt{\frac{60*0.9}{0.80} }\\ v = 8.215 m/s[/tex]

The maximum speed the stone can attain is 8.215 m/s

Learn more on centripetal force here;

https://brainly.com/question/7889464