Answer:
v(t) = 21.3t
v(t) = 5.3t
[tex]v(t) = 48 -48 e ^{ \frac{t}{9}}[/tex]
Explanation:
When no sliding friction and no air resistance occurs:
[tex]m\frac{dv}{dt} = mgsin \theta[/tex]
where;
[tex]\frac{dv}{dt} = gsin \theta , 0 < \theta < \frac{ \pi}{2}[/tex]
Taking m = 3 ; the differential equation is:
[tex]3 \frac{dv}{dt}= 128*\frac{1}{2}[/tex]
[tex]3 \frac{dv}{dt}= 64[/tex]
[tex]\frac{dv}{dt}= 21.3[/tex]
By Integration;
[tex]v(t) = 21.3 t + C[/tex]
since v(0) = 0 ; Then C = 0
v(t) = 21.3t
ii)
When there is sliding friction but no air resistance ;
[tex]m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta[/tex]
Taking m =3 ; the differential equation is;
[tex]3 \frac{dv}{dt}=128*\frac{1}{2} -\frac{\sqrt{3} }{4}*128*\frac{\sqrt{3} }{4}[/tex]
[tex]\frac{dv}{dt}= 5.3[/tex]
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 :
[tex]m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta - kv[/tex]
The differential equation is :
= [tex]3 \frac{dv}{dt}=128*\frac{1}{2} - \frac{ \sqrt{ 3}}{4}*128 *\frac{ \sqrt{ 3}}{2}-\frac{1}{3}v[/tex]
= [tex]3 \frac{dv}{dt}=16 -\frac{1}{3}v[/tex]
By integration
[tex]v(t) = 48 + Ce ^{\frac{t}{9}[/tex]
Since; V(0) = 0 ; Then C = -48
[tex]v(t) = 48 -48 e ^{ \frac{t}{9}}[/tex]
