Answer to Question #104273 in Mechanics | Relativity for Praveen

Question

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

Answer 1

$F=ma$

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

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

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