A bullet of mass 20kg travel ling horizontally with a Speed of 500 m/s passes through a wooden block of mass 10 kg initially at rest on the surface the bullet emerges with a speed of 100m/s and the block slide 20cm on the surface before coming to rest the coefficient of friction between the block and the surface.(g=10m/s2)?

First of all your question is a bit wrong. The mass of bullet will be 20g buddy.Look here were will have to conserve momentum in the horizontal direction.Total momentum initially = total momentum finally=> (0.02)(500) = 0.02×100 + 10 × v=> v = (10 - 2)/10 = 0.8m/sThe block had an initial velocity 0.8m/s and came to rest after 0.2m .W = change in mechanical energyLet coefficient of friction be zzmgx = 1/2 (mv^2)=> z = 0.5×0.8×0.8÷ (10×0.2)=> z = 0.16This seems an acceptable value for coefficient of friction but if you`d taken 20kg of bullet , it would have been unrralistic. Hope this is helpful.