An object of mass 0.675 kg on a frictionless table is attached to a string that

Question

An object of mass 0.675 kg on a frictionless table is attached to a string that passes through a hole in the table at the center of the horizontal circle in which the object moves with constant speed. If the radius of the circle is 0.500 m and the speed is 10.0 m/s, (a) compute the tension in the string. (b) It is found that drawing an additional 0.200 m of the string down through the hole, thereby reducing the radius of the circle to 0.300 m, has the effect of multiplying the original tension in the string by 4.63. c) Computer the total work done by the string on the revolving object during the reduction of the radius.
Any lifesavers out there? Completely lost on this one. Will be much appreciated!!

Explanation / Answer

a) tension = mv^2 / r = 0.675*10^2 / 0.500 = 135N b) new tension = 4.63 * 135 =625.05 new tension = mv^2 / r = 635.05 new v = 16.66m/sec c) work done = change in kinetic energy = 1/2 m(new v)^2 - 1/2 mv^2 = 59.92 J

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