A 2.5kg mass tied to a string is whirled around a horizontal circle of radius 1.7m. The tension in the string is 35.7N and the string is inclined at 60° above the horizontal. Calculate the speed at which the stone moves round the circle.
According to the second Newton's law "F=ma". So, we have
"N\\cos60\u00b0=m\\frac{v^2}{r} \\to v=\\sqrt{\\frac{N\\cos60\u00b0\\cdot r}{m}}="
"=\\sqrt{\\frac{35.7\\cdot\\cos60\u00b0\\cdot 1.7}{2.5}}\\approx3.5(m\/s)" . Answer
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