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## Homework Statement

There are two fixed inclined planes as shown (ABC and DGEF). AB=DE=y, BC=EF=x. The coeff of friction between the wedges and the mass m is [tex]\mu[/tex].

A small mass M is released at the point A and reaches the bottom with a speed V1. The same block is released from point D and reaches F with speed V2. Which is the correct option:

V1>V2

V1<V2

V1=V2.

## The Attempt at a Solution

I thought since the potential energy is same here in both cases, but the path length DGF is greater than AC, therefore more energy must be dissipated by friction and so V1>V2.

But the answer is V1=V2.

How?