Since Tina is sliding down on an inclined ramp
Net force along the inclined
[tex]F_x = mg sin\theta[/tex]
Now force component on Tina perpendicular to inclined plane
[tex]F_y = mg cos\theta[/tex]
now we know that normal to the inclined plane the force is counter balanced by the normal force
So we can find the normal force as
[tex]F_n = mg cos\theta[/tex]
now in order to find the friction force we can write
[tex]F_f = \mu * F_n[/tex]
[tex]F_f = 0.10*mgcos\theta[/tex]
now along the inclined plane net force is given as
[tex]F_{net} = mgsin\theta - F_f[/tex]
[tex]F_{net} = mgsin\theta - 0.10*mgcos\theta[/tex]
also by Newton's II law we can write
[tex]F_{net} = ma[/tex]
by above two equations we can write
[tex]ma = mgsin\theta - 0.10*mgcos\theta[/tex]
[tex]a = gsin\theta - 0.10*gcos\theta[/tex]
[tex]a = 9.81*sin30 - 0.10*9.81*cos30[/tex]
[tex]a = 4.06 m/s^2[/tex]
so acceleration will be 4.06 m/s^2