Answer:
Explanation:
distance covered by ice cube = 0.5 m
force acting downwards = mgsin49
force of friction acting along surface upwards
= mgcos49 x .16 ( reaction force of inclined surface R = mgcos49.)
net force acting downwards
mgsin49 - mgcos49 x .16
mg (sin49 - cos49 x .16 )
net acceleration downwards
g (sin49 - cos49 x .16 )
= (.7547 - .105 ) x 9.8
= 6.367 m / s²
s = ut + 1/2 at²
0.5 = 0 + 1/2 x 6.367 x t²
t² = .1570
t = .4 s
b )
Let required inclination be θ
in the second case net acceleration
a = g ( sin θ - .385 cosθ )
s = ut + 1/2 at²
0.5 = 0 + 1/2 x g ( sin θ - .385 cosθ ) x .4 x .4
( sin θ - .385 cosθ ) = .6377
sin θ = .385 cosθ + .6377
θ = 58 degree.