An ice cube with a mass of 0.0600 kg is placed at the midpoint of a 1.00-m-long wooden board that is propped up at a 49° angle. The coefficient of kinetic friction between the ice and the wood is 0.160.
(a) How much time does it take for the ice cube to slide to the lower end of the board?
(b) If the ice cube is replaced with a 0.0600 kg wooden block, where the coefficient of kinetic friction between the block and the board is 0.385, at what angle should the board be placed so that the block takes the same amount of time to slide to the lower end as the ice cube does?

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.

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