Answer:
[tex]v_f = 3.45 m/s[/tex]
Explanation:
As we know that box was initially at rest
so here work done by all forces on the box = change in its kinetic energy
so we will have
[tex]mg sin\theta L = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2[/tex]
now we have
m = 10 kg
[tex]\theta = 10^0[/tex]
L = 3.5 m
so we will have
[tex]10(9.81)sin10 \times 3.5 = \frac{1}{2}(10)(v_f^2 - 0)[/tex]
[tex]2(9.81) sin10 \times 3.5 = v_f^2[/tex]
so final speed is given as
[tex]v_f = 3.45 m/s[/tex]
