Respuesta :
Answer:1.89 m
Explanation:
Given
Block travels [tex]0.63\ m[/tex] in first second
It is released from rest i.e. initial speed is zero (u=0)
using
[tex]s=ut+\frac{1}{2}at^2[/tex]
where a=acceleration
here acceleration is the component of gravity on incline plane (say [tex]\theta [/tex])
so
[tex]s_1=\frac{1}{2}\times g\sin \theta (1)^2[/tex]
[tex]0.633\times 2=9.8\sin \theta \times 1^2[/tex]
[tex]\sin\theta =0.1291[/tex]
[tex]\theta =7.41^{\circ}[/tex]
So distance traveled in [tex]2\ sec[/tex]
[tex]s=\frac{1}{2}\times g\sin \theta (2)^2[/tex]
[tex]s=0.5\times 9.8\times \sin (7.41)\times 4[/tex]
[tex]s=2.52\ m[/tex]
So distance traveled in [tex]2^{nd}\ sec[/tex] is
[tex]s-s_1=2.52-0.633=1.89\ m[/tex]
