A box with an initial speed of #4 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #5/6 # and an incline of #pi /3 #. How far along the ramp will the box go?

1 Answer
Nov 10, 2016

Answer:

#s=v^2/(2g(sin alpha+ k cos alpha))#

#s=0.64m#

Explanation:

#E_k=E_p+W_T#
#mv^2/2=mgh+F_ts#
#F_t=F_n=mg*cos alpha#
#h/s=sin alpha#
#mv^2/2=mgs *sin alpha+kmgs cos alpha#
#s=v^2/(2g(sin alpha+ k cos alpha))#
#k=5/6#
#alpha=pi/3#
#g=9.81m/s^2#
#s=0.64m#