Recent content by amgoda

yup.. i just messed up in the first calculation

ahaaa .. here we go .. i don't know why i haven't thought of this Wnet=μk.mg.Δx = change in KE then KE at the bottom = 13.5 =mgh h= .027 and i recalculated H in the First method i used and i got 0.27 also .. thanks a lot u drew my attention to the easier way .. thanks

ahaa .. so it should equal to the kinetic energy at the end of the horizontal plane .. so will it be .5*m*v^2=mgh .5*5*2.32^2=5*9.8*h h=0.63 meters ?

that's the problem .. i just got to this point and i don't know what to do afterwards ... btw thanks for the fast replies

alright .. here is how i thought about it .. N=mg Fκ=μk.N a=(μk.mg)/-m=-.98m/s^s vf^2=V0+2a(x-x0) vf^2=5^2+(2*-.98*10)= 2.32 and i thought that the final velocity on the horizontal plane is the initial velocity for the inclined plane ..

1. A body with mass m = 5 kg and initial velocity = 5 m/s moves along a horizontal plane with length 10 m, where the friction coefficient is μ=0.1, After passing the part with friction, the body moves up along a tilted plane which has no friction. What is the maximum height of body elevation...