Recent content by Speags

a hockey puck of mass m is sliding in the +x direction across a horizontal ice surface. while sliding, the puck is subject to two forces that oppose its motion: a constant sliding friction force of magnitude f, and a air resistance force of magnitude cv^2 , where c is a constant and v is the...

i need to intergrate this function \frac{2mdx}{cx^2} from x to 0

i have a force acting on and object and I've gotten the question down to when i have to intergrate but I'm stuck on the intergral F=-(f+cv^2) m \frac{dv}{dx} \frac {dx}{dt} = -(f+cv^2) f and c are constants mvdv=-(f+cv^2)dx \frac{mvdv}{f+cv^2}=-dx now how do i intergrat them from...

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...

v_{o} = 0 , x_{o} = 0 thanks for your help

m \int_{v_{o}}^{v} vdv= F_{o} \int_{x_{o}}^{x} dx + C_{x} \int_{x_{o}}^{x} dx should be m \int_{v_{o}}^{v} vdv= F_{o} \int_{x_{o}}^{x} dx + C \int_{x_{o}}^{x} xdx sorry i didn't explain it well enough C and X are seperate so does that mean the solution would be: 1/2mv^2=F_{o}x...

the question asks, Find the velocity v as a function of the displacement x for a particle of mass m, which starts from rest x=0 F(x)=Fo + Cx where Fo and C are positive constants So far I've gotten, ma=Fo + Cx m (dv/dt)=Fo +Cx m (dv/dx dx/dt)=Fo +Cx I...