Answer:
[tex]2.4\sqrt{L}[/tex] where L is the length of the ramp
Explanation:
Let L (m) be the length of the ramp, and g = 9.81 m/s2 be the gravitational acceleration acting downward. This g vector can be split into 2 components: parallel and perpendicular to the ramp.
The parallel component would have a magnitude of
[tex]gsin\theta = 9.81 sin17^o = 2.87 m/s^2[/tex]
We can use the following equation of motion to find out the final velocity of the book after sliding L m:
[tex]v^2 - v_0^2 = 2a\Delta s[/tex]
where v m/s is the final velocity, [tex]v_0[/tex] = 0m/s is the initial velocity when it starts from rest, a = 2.87 m/s2 is the acceleration, and [tex]\Delta s = L[/tex] is the distance traveled:
[tex]v^2 - 0 = 2*2.87*L[/tex]
[tex]v = \sqrt{5.74L} = 2.4\sqrt{L}[/tex]
