Question

In the figures (a) and (b); AC, DG and GF are fixed inclined planes with AB = DE = h. BC = EF = x and EL = LF = x/2. A small block of mass M is released from A which slides down AC and reaches C with a velocity v₁ . The same block is now released from D. It slides down along DGF reaching F with a velocity v₂ . The coefficients of friction between the block and surfaces AC; DG and GF are μ; μ/2 and μ/4 as shown. The ratio v₂/v₁ is 

(1) \([\frac{8h-3μx}{8(h-μx)}]^{\frac{1}{2}}\)

(2) \([\frac{8h-8μx}{8(h-μx)}]^{\frac{1}{2}}\)

(3) \(\sqrt{6\frac{h}{x}}\)

(4) \(\sqrt{\frac{x}{6h}}\)

Answer 1

 (1) \([\frac{8h-3μx}{8(h-μx)}]^{\frac{1}{2}}\)

For Fig. (a); the KE at C is the difference of PE lost and the work against friction.

From eqns (1) and (2) we have