## 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

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.)

## Answers ( )

Answer:v(t) = 21.3tv(t) = 5.3tExplanation:When no sliding friction and no air resistance occurs:

where;

Taking m = 3 ; the differential equation is:

By Integration;

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

v(t) = 21.3tii)When there is sliding friction but no air resistance ;

Taking m =3 ; the differential equation is;

By integration; we have ;

v(t) = 5.3tiii)To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

The differential equation is :

=

=

By integration

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