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

Answer:

v(t) = 21.3t

v(t) = 5.3t

$v(t) = 48 -48 e ^{ \frac{t}{9}}$

Explanation:

When no sliding friction and no air resistance occurs:

$m\frac{dv}{dt} = mgsin \theta$

where;

$\frac{dv}{dt} = gsin \theta , 0 < \theta < \frac{ \pi}{2}$

Taking m = 3 ; the differential equation is:

$3 \frac{dv}{dt}= 128*\frac{1}{2}$

$3 \frac{dv}{dt}= 64$

$\frac{dv}{dt}= 21.3$

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 ;

$m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta$

Taking m =3 ; the differential equation is;

$3 \frac{dv}{dt}=128*\frac{1}{2} -\frac{\sqrt{3} }{4}*128*\frac{\sqrt{3} }{4}$

$\frac{dv}{dt}= 5.3$

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 :

$m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta - kv$

The differential equation is :

= $3 \frac{dv}{dt}=128*\frac{1}{2} - \frac{ \sqrt{ 3}}{4}*128 *\frac{ \sqrt{ 3}}{2}-\frac{1}{3}v$

= $3 \frac{dv}{dt}=16 -\frac{1}{3}v$

By integration

$v(t) = 48 + Ce ^{\frac{t}{9}$

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

$v(t) = 48 -48 e ^{ \frac{t}{9}}$