Question

In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)

Answer 1

v(t) = 21.3t

v(t) = 5.3t

`v(t) = 48 -48 e ^{ (t)/(9)}`

Step-by-step explanation:

When no sliding friction and no air resistance occurs:

`m(dv)/(dt) = mgsin \theta`

where;

`(dv)/(dt) = gsin \theta , 0 < \theta < ( \pi)/(2)`

Taking m = 3 ; the differential equation is:

`3 (dv)/(dt)= 128*(1)/(2)`

`3 (dv)/(dt)= 64`

`(dv)/(dt)= 21.3`

By Integration;

`v(t) = 21.3 t + C`

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

`m (dv)/(dt)= mg sin \theta - \mu mg cos \theta`

Taking m =3 ; the differential equation is;

`3 (dv)/(dt)=128*(1)/(2) -(√(3) )/(4)*128*(√(3) )/(4)`

`(dv)/(dt)= 5.3`

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

`m (dv)/(dt)= mg sin \theta - \mu mg cos \theta - kv`

The differential equation is :

=
`3 (dv)/(dt)=128*(1)/(2) - ( √( 3))/(4)*128 *( √( 3))/(2)-(1)/(3)v`

=
`3 (dv)/(dt)=16 -(1)/(3)v`

By integration

`v(t) = 48 + Ce ^{(t)/(9)`

Since; V(0) = 0 ; Then C = -48

`v(t) = 48 -48 e ^{ (t)/(9)}`