(4) An object with a mass m slides across a frictionless horizontal surface. At

Question

(4) An object with a mass m slides across a frictionless horizontal surface. At time t 0 it has an initial velocity v-vo and its initial position is x = 0, A drag force Fa acts on the object and depends both linearly and quadratically on the velocity: where k1 and k2 are constant coefficients. (a) Neglecting the gravitational force, find the instantaneous velocity, v(t), and position, (t), of the object. (b) Will the object eventually come to rest at a finite distance from where it enters? If so where does it stop? (over)

Explanation / Answer

given opbject of mass m

sliding on frictionless horizontal surface

at time t, initial position x = 0, initial speed = vo

drag force Fd = -k1v - k2v^2

a. from newtons second law

m dv/dt = -k1v - k2v^2

dv/v(k1 + k2v) = -dt/m

consider

1/v(k1 + k2v) = A/v + B/(k1 + k2v)

then

A(k1 + k2v) + Bv = 1

for v = 0

A = 1/k1

for v = -k1/k2

B = -k2/k1

hence

(1/v - k2/(k1 + k2v))dv = -k1*dt/m

integrasting both sides

ln(k1/v + k2) = k1*t/m + C

at t = 0, v = vo

ln(k1/vo + k2) = c

hence

ln[(k1/v + k2)/(k1/vo + k2)] = k1*t/m

v = k1/[(k1/vo + k2)e^(k1*t/m) + k2 ]

now dx/dt = v = k1/[(k1/vo + k2)e^(k1*t/m) + k2 ]

dx = k1*dt/[(k1/vo + k2)e^(k1*t/m) + k2 ]

let (k1/vo + k2)e^(k1*t/m) + k2 = g

then e^*k1*t/m = (g - k2)/(k1/vo + k2)

(k1/vo + k2)k1*e^(k1*t/m)dt/m = dg

k1* (g - k2)dt/m = dg

dt = m*dg/(g - k2)

hence

dx/k1*m = dg/(g - k2)*g

consider 1/(g - k2)g = A/g + B/(g - k2)

A(g - k2) + Bg = 1

when g = 0

A = -1/k2

when g = k2

B = 1/k2

hence

k2*dx/k1*m = (-1/g + 1/(g - k2))dg

integrating

k2*x/k1*m = ln(1 - k2/g) = ln(1 - k2/[(k1/vo + k2)e^(k1*t/m) + k2]) + K

at t = 0, x = 0

0 = ln(1 - k2/g) = ln(1 - k2/[(k1/vo + 2k2]) + K

K = -ln(1 - k2/[(l1/vo + 2k2)])

hence

x = (m*k1/k2)ln(1 - k2/[(k1/vo + k2)e^(k1*t/m) + k2]) - -ln(1 - k2/[(l1/vo + 2k2)])

now say v = 0 at t = t

v = k1/[(k1/vo + k2)e^(k1*t/m) + k2 ] = 0

this will never happen as k1 != 0

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