A box with an initial speed of #1 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #2/3 # and an incline of #(3 pi )/8 #. How far along the ramp will the box go?
1 Answer
I would take the sum of the parallel and perpendicular forces. The perpendicular forces (
As the box moves forward, the kinetic friction force resists it, and the parallel component of the gravitational force resists it too. That means both those forces are negatively-signed with respect to forward motion.
So:
#sum vecF_(||) = -vecF_k - vecF_(g,||)#
#= -mu_kvecF_N - mvecgsintheta = mveca_(||)# where we have put the signs in the equation,
#mu_k = 2/3# , and#vecg > 0# .
#sum vecF_(_|_) = vecF_N - vecF_(g,_|_) = vecF_N - mvecgcostheta = 0#
As a result,
#vecF_N = mvecgcostheta#
and
#-mu_kcancel(m)vecgcostheta - cancel(m)vecgsintheta = cancel(m)veca_(||)#
Therefore:
#veca_(||) = -(mu_k vecg cos theta + vecg sin theta)#
#= -(2/3 cdot "9.81 m/s"^2 cdot cos((3pi)/8) + "9.81 m/s"^2 cdot sin((3pi)/8))#
#= -"11.57 m/s"^2#
This says that logically, the box will slow down to zero velocity. Since the ramp is straight, we can assume this is the average acceleration:
#veca_(||) -= (Deltavecv_(||))/(Deltax)#
Letting the bottom of the ramp be the initial position
#veca_(||) = (0 - vecv_i)/(x_f - 0)#
As a result,
#color(blue)(x_f) = -(vecv_i)/(veca_(||)) = -("1 m/s")/(-"11.57 m/s"^2)#
#=# #color(blue)"0.086 m"# up the ramp.
That's about
