A 0.2-kg stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface.If the stone makes 150 revolutions per minute, the tension force of the string on the stone is:
A)0.03 N
B)0.2 N
C)0.75 N
D)1.96 N
E)30 N

Question

Quizplus · Accepted Answer

The tension force is the centripetal force required to keep the stone moving in a circle. First, convert revolutions per minute to radians per second: 150 rev/min = 150 * 2π rad/60 s = 5π rad/s. The centripetal force is given by F = m * ω² * r, where m = 0.2 kg, ω = 5π rad/s, and r = 0.6 m. Calculating gives F = 0.2 * (5π)² * 0.6 = 30 N.

A 02-Kg Stone Is Attached to a String and Swung in in a Circle