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

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.)

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Thu Nguyệt 2 weeks 2021-07-20T04:46:11+00:00 1 Answers 2 views 0

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    2021-07-20T04:48:00+00:00

    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}}

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