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Blakemiller2020
06.05.2020 •
Physics
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.)
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Ответ:
v(t) = 21.3t
v(t) = 5.3t
Explanation:
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) = 21.3t
ii)
When there is sliding friction but no air resistance ;
Taking m =3 ; the differential equation is;
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 :
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](/tpl/images/0646/4201/22c8f.png)
=![3 \frac{dv}{dt}=16 -\frac{1}{3}v](/tpl/images/0646/4201/9a255.png)
By integration
Since; V(0) = 0 ; Then C = -48
Ответ: