Work, energy with kinetic friction

1. Nov 2, 2004

Bcisewski

Can anyone provide some assistance? I know this comes in two parts, one in locating the velocity, which I belive comes down to v=sq root of 2(9.8)(6.34), however the second part is creating havoc. Any suggestions on part II's formula?

A box slides down a frictionless 6.34 m high hill, starting from rest. At the bottom of the hill, the box slides along a level surface where the coefficient of kinetic friction uk = 0.246. How far from the bottom of the hill does the box come to rest? The final answer will be 25.8m

2. Nov 2, 2004

Pyrrhus

The First part

$$mgh = \frac{1}{2}mv^2$$

The Second Part:

1)

Use the kinetic constant acceleration formula

$$v^2 = v_{o}^2 + 2a(x-x_{o})$$

and Newton's 2nd Law on the box where there's friction to find the acceleration

$$\sum_{i=1}^{n} \vec{F}_{i} = m \vec{a}$$

2)

Use

$$\Delta E = W_{f}$$

$$\frac{1}{2}mv^2 = \mu mgd$$

Last edited: Nov 2, 2004