Abhimanyu`` Zala
Last Activity: 13 Years ago
As there is no mention of friction the block will undergo translational motion with constant accelearation of gsinβ (note βis the angle of inlclination of the inclined plane wit the ground).
Using,
S=ut+0.5at2
S=0.5at2 (initial velocity = 0)
0.5=0.5a(0.5)2
0.5=0.5(10sinβ)0.25 (g=10)
sinβ=2/5
Now as the half second is elapsed it requires the final velocity at the end of first half second is given by
v=u+at
v=gsinβt (u=0)
v=(10)(2/5)(1/2)
v=2m/s
But the above same velocity becomes the initial velocity for the second 1/2 second so now,
S=ut+0.5at2
(1/2)=(2)t+0.5(10x2/5)t2
We Get,
A quadratic equation,
4t2 + 4t + -1 = 0
Solving we get
t=0.21 seconds
ANSWER:-Thus the block will require more 0.21 seconds to cover the more half of the distance.