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

1. Jun 21, 2006

### 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 $\mu_k\,=\,0.6$.

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?

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

Last edited: Jun 21, 2006
2. Jun 21, 2006

### Hootenanny

Staff Emeritus
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.

Last edited: Jun 21, 2006
3. Jun 21, 2006

### 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?

4. Jun 21, 2006

### Hootenanny

Staff Emeritus
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?

Last edited: Jun 21, 2006