A block is released from the top of two inclined rough surfaces of height h each. The angle of inclination of the two planes are `30^(@)` and `60^(@)` respectively. All other factors (e.g., coefficient of friction, mass of block etc.) are same in both the cases. Let `K_(1)` and `K_(2)` be the kinetic energies of the block at the bottom of the plane in two cases. Then,

Options

A. `K_1 = K_2`
B. `K_1 gt K_2`
C. `K_1 lt K_2`
D. Data sufficient

Answer

C

Explanation

loss in GPE = gain KE + work done against friction.
`implies mgh = K + mg cos theta. h/(sin theta) (mu) " "implies " " K = mgh(1 - mu cot theta)`
`implies K_1 = mgh (1 - mu sqrt(3)) and K_2 = mgh (1-(mu)/(sqrt3)) implies K_1 < K_2` .