# Horizontal motion with quadratic resistance

1. Oct 12, 2004

### Speags

so i got a block with mass=m traveling on an oiled surface. the block suffers a viscous resistance given:
$$F(v)= -cv^{3/2}$$
the initial speed of the block is $$v_{o}$$ at x=0, i have to show that the block cannot travel farther than $$2mv_{o}^{1/2} /c$$
so far i have;
$$ma=-cv^{3/2}$$
$$m \frac{dv}{dx} \frac{dx}{dt} = -cv^{3/2}$$
$$mvdv=-cv^{3/2} dx$$
$$dx= \frac {mvdv}{cv^{3/2}}$$

where should i go from here?

2. Oct 12, 2004

### ehild

Put back the "-" sign you have lost,

simplify

$$dx= \frac {-mdv}{cv^{1/2}}$$

Integrate from x=0 to x(final) and from v=v0 to v=0 and you get the desired result.

ehild