A body of mass m is released from the top of a rough inclined plane. If the frictional force be fk. Then prove that the body will reach the bottom with velocity given by v= under root 2/m(mgh - fL)..{want a step wise answer}

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bullet of mass 0.02 kg traveling horizontally with velocity 250ms-1 strikes a block of wood of mass 0.23 kg which rests on a rough horizontal surface. After the impact, the block and bullet move together and come to rest after traveling a distance of 40m. The coefficient of sliding friction of the rough surface is : (g = 9.8ms-2) (2009 E) 1) 0.75 2) 0.61 3) 0.51 4) 0.30 Ans : 3 Sol: After the collision the bullet and block move combinedly and comes to rest after covering a distance of 40m. From law of conservation of linear momentum mu mu mv m v 11 2 2 11 2 2 + =+ ⇒ ×+ × 0.02 250 0.23 0= (0.02 + 0.23) v ⇒ 500 1 20 25 v ms− = = From law of conservation of energy ( ) 1 2 2 ⇒ = Mv Mg S μ 1 0.25 400 2 ⇒× × = μ ×0.25 9.8 40 × × ⇒ = μ 0.51