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A partical is realeased from the top of ...

A partical is realeased from the top of two inclined rought surface 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 the block etc) are same in both the cases. Let `K_(1) and K_(2)` be the kinetic energy of the partical at the bottom of the plane in two cases. Then

A

`K_1 = K_2`

B

`K_1 gt K_2`

C

`K_1 lt K_2`

D

Data sufficient

Text Solution

Verified by Experts

The correct Answer is:
C
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` .
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