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Speags
Oct12-04, 01:53 PM
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?

ehild
Oct12-04, 02:16 PM
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;
.....

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

where should i go from here?

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