A .275 kg object is swung in a vertical circular path ona string .850 m long. If

Question

A .275 kg object is swung in a vertical circular path ona string .850 m long. If the speed is 5.06 m/s at the top of the circle, what is thetension of the string there?________N If the string breaks when its tension exceeds 21.5 N, what isthe maximun speed the object can have at the bottom before the stringbreaks?__________m/s A .275 kg object is swung in a vertical circular path ona string .850 m long. If the speed is 5.06 m/s at the top of the circle, what is thetension of the string there?________N If the string breaks when its tension exceeds 21.5 N, what isthe maximun speed the object can have at the bottom before the stringbreaks?__________m/s

Explanation / Answer

at the top, let the tension be T, we have: T+mg = ma = mv2 / r =>T = mv2/r-mg=m[v2/r-g] =0.275kg*[(5.06m/s)2/0.850m - 9.8m/s2] =5.59N at the top of the circle, the tension of the rope is5.59N at the bottom of the circle, we have: T-mg = mv2/r here T = 21.5N, =>v = [Tr/m - rg] = [21.5N*0.850m/0.275kg -0.850m*9.8m/s2] = 7.62 m/s the maximun speed is v = 7.62m/s

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