A box with an initial speed of #6 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #1/2 # and an incline of #(3 pi )/8 #. How far along the ramp will the box go?

1 Answer
Jun 10, 2016

Answer:

I found #1.6m# (but double check my maths!)

Explanation:

Consider the diagram:
enter image source here
We can use Newton's Second Law: #SigmavecF=mveca#
Where:
along the horizontal (along the ramp):
#-mgsin(3pi/8)-f=ma#
along the vertical:
#N-mgcos(5pi/8)=0#
#f=muN# is friction
#mgsin(5pi/8)# and #mgcos(5pi/8)# are the horizontal and vertical components of weight.

We also need to change #a# into an expression involving the distance #s#;
we can use kinematics and:
#v_f^2=v_i^2+2as#
#a=(v_f^2-v_i^2)/(2s)#
with
#v_i=6m/s#
#v_f=0#

So we get:
#-mgsin(3pi/8)-f=ma#
#-mgsin(3pi/8)-muN=m(v_f^2-v_i^2)/(2s)#
#-cancel(m)gsin(3pi/8)-cancel(m)mugcos(5pi/8)=cancel(m)(v_f^2-v_i^2)/(2s)#
#-9.8sin(3pi/8)-(1/2)9.8cos(5pi/8)=(0^2-6^2)/(2s)#
#-9.1-1.8=-36/(2s)#
#s=1/2(36/10.9)=1.6m#