Solve Distance for Box A Sliding Down Ramp

[VinnyCee]

The 30 lb box A is released from rest and slides down along the smooth ramp and onto the surface. Determine the distance s from the end of the surface to where the box stops. The coefficient of kinetic friction between the cart and the box is \mu_k\,=\,0.6.

http://img224.imageshack.us/img224/8509/problem14343jq.jpg


Here is what I have so far:


-W\,\Delta\,y\,=\,(-30\,lb)\,(-4\,ft)\,=\,120\,ft\,lb

\sum\,F_y\,=\,N\,-\,W\,=\,0\,\Rightarrow\,N\,=\,W\,=\,30\,lb

\sum\,F_x\,=\,-f_k\,=\,m\,a_x\,\Rightarrow\,-\mu_k\,N\,=\,m\,a_x

(-0.6)\,(30\,lb)\,=\,(0.932)\,a_x

a_x\,=\,\frac{-18.6}{0.932}\,=\,-19.3\,\frac{ft}{s^2}

Now what?

I know I need to find v_f and the bottom of the hill and I am probably supposed to use a work-energy equation?

\sum\,T_1\,+\,\sum\,U_{1\,-\,2}\,=\,\sum\,T_2

Please help, thanks.

 

[Hootenanny]

know I need to find v_f and the bottom of the hill and I am probably supposed to use a work-energy equation?

Yes, as the ramp is frictionless the kinetic energy gained by the block will equal the work done by gravity; ¹/₂mv² = mgh. A good point to note for future reference is that the velocity of the object is independent of the mass.

 

[VinnyCee]

Using that, I get this:

V_f^2\,=\,2\,g\,h\,=\,2\,(32.2)\,(4)\,=\,257.6

V_f\,=\,\sqrt{257.6}\,=\,16.05\,\frac{ft}{s}

v\,=\,v_0\,+\,a\,t

0\,=\,16.05\,+(-19.3)\,t

t\,=\,0.832\,s

s\,=\,s_0\,+\,v_0\,t\,+\,\frac{1}{2}\,a\,t^2

s\,=\,0\,+\,(16.05)\,(0.832)\,+\,\frac{1}{2}\,a\,t^2

s\,=\,6.67\,ft

The real answer is 3.33 ft though! What did I do wrong?

 

[Hootenanny]

Your going to kick yourself for this one. You have calculated the distance travelled, not the distance from the end of the platform. HINT: What does 10 - 6.67 equal?