Answer to Question #202047 in Mechanics | Relativity for AYUSH VERMA

Question

A child of mass 30Kg skates down from the top of a ramp having a constant slope of 15degree. The child's speed increases from 1.5 metre per second to 3.0 metre per second as she reaches the bottom of the ramp. A force of kinetic friction of magnitude 50 Newton opposes her motion draw the appropriate force diagram and determine the length of the ramp take g=10meter per second square

Answer 1

Solution.

$m=30kg;$

$\alpha=15^o;$

$v_0=1.5 m/s;$

$v=3 m/s;$

$g=10 m/s^2;$

$F_f=50N;$

$ma=mgsin\alpha-F_f \implies a=\dfrac{mgsin\alpha-F_f}{m};$

$a=\dfrac{30kg\sdot10 m/s^2.0.2588-50N}{30kg}=0.92 m/s^2;$

$l=\dfrac{v^2-v_0^2}{2a};$

$l=\dfrac{9m^2/s^2-2.25m^2/s^2}{2\sdot0.92m/s^2}=3.67m;$

Answer: $l=3.67m.$