Horizontal motion with quadratic resistance

[Speags]

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

where should i go from here?

 

[ehild]

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

[tex] dx= \frac {mvdv}{cv^{3/2}} [/tex]

where should i go from here?

Put back the "-" sign you have lost,

simplify

[tex] dx= \frac {-mdv}{cv^{1/2}} [/tex]

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

ehild

 

[wais]

First, we can rewrite the equation to get rid of the variables dx and dt:

dx = \frac {mvdv}{cv^{3/2}}
dx = \frac {mv}{cv^{3/2}}dv

Next, we can integrate both sides of the equation from x=0 to x=d (where d is the distance traveled by the block):

\int_{0}^{d} dx = \int_{0}^{d} \frac {mv}{cv^{3/2}} dv

Using the power rule for integration, we get:

d = \frac {2mv^{1/2}}{c}

Now, we know that the maximum distance traveled by the block is when the final velocity (v) is equal to zero. So we can substitute v=0 into the equation:

d = \frac {2mv^{1/2}}{c}
d = \frac {2m(0)^{1/2}}{c}
d = 0

This means that the maximum distance traveled by the block is zero, which makes sense since the block cannot travel any further once it comes to a complete stop.

To prove that the block cannot travel farther than 2mv_{o}^{1/2} /c, we can substitute v_{o} for v in the equation:

d = \frac {2mv_{o}^{1/2}}{c}

This gives us the maximum distance traveled by the block, which is 2mv_{o}^{1/2} /c. Therefore, the block cannot travel any farther than this distance due to the quadratic resistance acting against it.