Question #104273

A child of mass 30 kg States down from the top of the ramphaving a constant slope of 15 degree the child's speed increased from 1.5 metre per second to 3.0 metre per second as series the bottom of the ramp a force of kinetic friction of magnitude 15 Newton opposes the motion draw the appropriate force dragon and determine the length of the ramp take g is equal to 10 metre per second square

Expert's answer

"F=ma"

"mg\\sin 15\u00b0-F_{fr}=ma \\to a=\\frac{mg\\sin 15\u00b0-F_{fr}}{m}=g\\sin 15\u00b0-\\frac{F_{fr}}{m}="

"=10\\cdot \\sin15\u00b0-\\frac{15}{30}\\approx2.1 \\frac{m}{s^2}"

"L=\\frac{v^2-v^2_0}{2a}=\\frac{3^2-1.5^2}{2\\cdot 2.1}\\approx1.61m" **Answer**

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