A 10.0 kg block is sliding at 8.94 m/s in a straight line across a horizontal, frictionless surface. The block then passes over a 4.00 m stretch of surface with friction. The kinetic coefficient of friction of this patch is .600. What is the velocity of the block after it passes over the section of surface with friction?

Question

shamim · Answer

Frictional force=coefficient of friction*m*g=.6*10*9.8=?

shamim · Answer

Now u hv to use the formula
F=ma
.6*10*9.8=10*a
a=?

shamim · Answer

Now u hv to use 
v=u+at
v=?

shamim · Answer

Feel free to ask for more clarification

anonymous · Answer

what do you plug in for u and t?

shamim · Answer

Sorry, u hv to use v^2=u^2-2*a*s
Put u=8.94m/s
s=4m
a=.6*9.8m/s^2

shamim · Answer

v=?

anonymous · Answer

got it thanks a bunch!!

shamim · Answer

Glad to help u!!!!