a-hockey-puck-of-mass-m-slides-on-a-flat-table-lubricated-with-petroleum-in-the-

Question

A hockey puck of mass M slides on a flat table lubricated with petroleum in the horizontal direction, exerting a drag force proportional to the cubic root of velocity, i.e. F_Dr = -cv^1/3 on the puck. If at t = 0, we know that v = v', find x and v as functions of time.

Accepted Answer

$v=\left(\left(v^{\prime}\right)^{\frac23}-\frac{2ct}{3M}\right)^{\frac32};$ $x=\left(-\frac{3M}{5c}\right)\left\lbrack\left(\left(v^{\prime}\right)^{\frac23}-\left(\frac{2ct}{3M}\right)\right)^{\frac52}-\left(v^{\prime}\right)^{\frac53}\right\rbrack$

Answer

$v=\left(\left(v^{\prime}\right)^{\frac23}-\frac{2ct}{3M}\right)^{\frac32};$ $x=\left(-\frac{3M}{5c}\right)\left\lbrack\left(\left(v^{\prime}\right)^{\frac23}-\left(\frac{2ct}{3M}\right)\right)^{\frac52}-\left(v^{\prime}\right)^{\frac53}\right\rbrack$

Answer

$v=\left(\left(v^{\prime}\right)^{\frac23}-\frac{2ct}{3M}\right)^{\frac32};$ $x=\left(-\frac{3M}{5c}\right)\left\lbrack\left(\left(v^{\prime}\right)^{\frac23}-\left(\frac{2ct}{3M}\right)\right)^{\frac52}-\left(v^{\prime}\right)^{\frac53}\right\rbrack$

Answer

$v=\left(\left(v^{\prime}\right)^{\frac23}-\frac{2ct}{3M}\right)^{\frac32};$ $x=\left(-\frac{3M}{5c}\right)\left\lbrack\left(\left(v^{\prime}\right)^{\frac23}-\left(\frac{2ct}{3M}\right)\right)^{\frac52}-\left(v^{\prime}\right)^{\frac53}\right\rbrack$

A hockey puck of mass M slides on a flat table lubricated with petroleum in the horizontal direction, exerting a drag force proportional to the cubic root of velocity, i.e. FDr = -cv1/3 on the puck. If at t = 0, we know that v = v', find x and v as functions of time.

A hockey puck of mass M slides on a flat table lubricated with petroleum in the horizontal direction, exerting a drag force proportional to the cubic root of velocity, i.e. F_Dr = -cv^1/3 on the puck. If at t = 0, we know that v = v', find x and v as functions of time.