## Block slides 4ft down a smooth plane & then slides on a rough surface (0.6) - s = ?

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 $\mu_k\,=\,0.6$.

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?

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

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 Quote by VinnyCee know I need to find vf 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.

 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?

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