Differential equations - mechanics question explanation

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I have a mechanics question that you have to solve with differential equations:




A particle of mass $0.4kg$ is moving at a speed of $10ms^-1$ when it enters an viscous liquid at a point, B. Inside the liquid the resistance force is proportional to the velocity, and initially equal to $1.2N$. Apart from the resistance an the weight, no other forces are acting on the particle.



Part A) Show that the velocity of the particle satisfies the differential equation $fracdvdt=-0.3v$.




So I got the answer wrong, of course.



My working out is:



$R:=kv$



$1.2:=10k$



$0.12:=k$



By $F=ma,$



$0.4g-0.12v:=:0.4fracdvdt$



So $g-0.3v = fracdvdt$



I don't understand how they managed to eliminate g? Or is my whole process wrong? What am I missing or fail to infer?




differential-equations proof-verification classical-mechanics




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  • 1




    But what about the force of weight? The question says weight is acting on the particle.
    – Cheks Nweze
    2 days ago










  • The viscous or otherwise called the resistance force is the only x component force, the weight is a y component equated to the bouyancy of the liquid. Thus your viscous force acts in the opposite direction to decelerate the mass, thus you have $mfracdvdt = -0.12v$ giving you $fracdvdt = -0.3v$
    – Satish Ramanathan
    2 days ago















up vote
0
down vote

favorite












I have a mechanics question that you have to solve with differential equations:




A particle of mass $0.4kg$ is moving at a speed of $10ms^-1$ when it enters an viscous liquid at a point, B. Inside the liquid the resistance force is proportional to the velocity, and initially equal to $1.2N$. Apart from the resistance an the weight, no other forces are acting on the particle.



Part A) Show that the velocity of the particle satisfies the differential equation $fracdvdt=-0.3v$.




So I got the answer wrong, of course.



My working out is:



$R:=kv$



$1.2:=10k$



$0.12:=k$



By $F=ma,$



$0.4g-0.12v:=:0.4fracdvdt$



So $g-0.3v = fracdvdt$



I don't understand how they managed to eliminate g? Or is my whole process wrong? What am I missing or fail to infer?




differential-equations proof-verification classical-mechanics




share|cite|improve this question















  • 1




    But what about the force of weight? The question says weight is acting on the particle.
    – Cheks Nweze
    2 days ago










  • The viscous or otherwise called the resistance force is the only x component force, the weight is a y component equated to the bouyancy of the liquid. Thus your viscous force acts in the opposite direction to decelerate the mass, thus you have $mfracdvdt = -0.12v$ giving you $fracdvdt = -0.3v$
    – Satish Ramanathan
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have a mechanics question that you have to solve with differential equations:




A particle of mass $0.4kg$ is moving at a speed of $10ms^-1$ when it enters an viscous liquid at a point, B. Inside the liquid the resistance force is proportional to the velocity, and initially equal to $1.2N$. Apart from the resistance an the weight, no other forces are acting on the particle.



Part A) Show that the velocity of the particle satisfies the differential equation $fracdvdt=-0.3v$.




So I got the answer wrong, of course.



My working out is:



$R:=kv$



$1.2:=10k$



$0.12:=k$



By $F=ma,$



$0.4g-0.12v:=:0.4fracdvdt$



So $g-0.3v = fracdvdt$



I don't understand how they managed to eliminate g? Or is my whole process wrong? What am I missing or fail to infer?




differential-equations proof-verification classical-mechanics




share|cite|improve this question











I have a mechanics question that you have to solve with differential equations:




A particle of mass $0.4kg$ is moving at a speed of $10ms^-1$ when it enters an viscous liquid at a point, B. Inside the liquid the resistance force is proportional to the velocity, and initially equal to $1.2N$. Apart from the resistance an the weight, no other forces are acting on the particle.



Part A) Show that the velocity of the particle satisfies the differential equation $fracdvdt=-0.3v$.




So I got the answer wrong, of course.



My working out is:



$R:=kv$



$1.2:=10k$



$0.12:=k$



By $F=ma,$



$0.4g-0.12v:=:0.4fracdvdt$



So $g-0.3v = fracdvdt$



I don't understand how they managed to eliminate g? Or is my whole process wrong? What am I missing or fail to infer?





differential-equations proof-verification classical-mechanics





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked 2 days ago









Cheks Nweze

466




466







  • 1




    But what about the force of weight? The question says weight is acting on the particle.
    – Cheks Nweze
    2 days ago










  • The viscous or otherwise called the resistance force is the only x component force, the weight is a y component equated to the bouyancy of the liquid. Thus your viscous force acts in the opposite direction to decelerate the mass, thus you have $mfracdvdt = -0.12v$ giving you $fracdvdt = -0.3v$
    – Satish Ramanathan
    2 days ago













  • 1




    But what about the force of weight? The question says weight is acting on the particle.
    – Cheks Nweze
    2 days ago










  • The viscous or otherwise called the resistance force is the only x component force, the weight is a y component equated to the bouyancy of the liquid. Thus your viscous force acts in the opposite direction to decelerate the mass, thus you have $mfracdvdt = -0.12v$ giving you $fracdvdt = -0.3v$
    – Satish Ramanathan
    2 days ago








1




1




But what about the force of weight? The question says weight is acting on the particle.
– Cheks Nweze
2 days ago




But what about the force of weight? The question says weight is acting on the particle.
– Cheks Nweze
2 days ago












The viscous or otherwise called the resistance force is the only x component force, the weight is a y component equated to the bouyancy of the liquid. Thus your viscous force acts in the opposite direction to decelerate the mass, thus you have $mfracdvdt = -0.12v$ giving you $fracdvdt = -0.3v$
– Satish Ramanathan
2 days ago





The viscous or otherwise called the resistance force is the only x component force, the weight is a y component equated to the bouyancy of the liquid. Thus your viscous force acts in the opposite direction to decelerate the mass, thus you have $mfracdvdt = -0.12v$ giving you $fracdvdt = -0.3v$
– Satish Ramanathan
2 days ago











1 Answer
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To have a compatible answer with the proposed question we have to consider the movement as horizontal.



After entering horizontally, the horizontal dynamics are described as



$$
m ddot x + kdot x = 0
$$



or



$$
mdot v_x + k v_x = 0
$$



solving this we obtain



$$
v_x = v_x^0 e^-frac km t
$$



but according to the initial conditions $v_x^0 = 10$ and also



$$
-F_0 = m dot v_x(0) = -m frac km v_x^0Rightarrow 1.2 = k v_x^0 = k times 10
$$



so we have



$$
k = 0.12Rightarrow frac km = 0.3
$$



hence the horizontal dynamics are dictated by



$$
dot v_x = -0.3 v_x
$$






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

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    up vote
    0
    down vote













    To have a compatible answer with the proposed question we have to consider the movement as horizontal.



    After entering horizontally, the horizontal dynamics are described as



    $$
    m ddot x + kdot x = 0
    $$



    or



    $$
    mdot v_x + k v_x = 0
    $$



    solving this we obtain



    $$
    v_x = v_x^0 e^-frac km t
    $$



    but according to the initial conditions $v_x^0 = 10$ and also



    $$
    -F_0 = m dot v_x(0) = -m frac km v_x^0Rightarrow 1.2 = k v_x^0 = k times 10
    $$



    so we have



    $$
    k = 0.12Rightarrow frac km = 0.3
    $$



    hence the horizontal dynamics are dictated by



    $$
    dot v_x = -0.3 v_x
    $$






    share|cite|improve this answer



























      up vote
      0
      down vote













      To have a compatible answer with the proposed question we have to consider the movement as horizontal.



      After entering horizontally, the horizontal dynamics are described as



      $$
      m ddot x + kdot x = 0
      $$



      or



      $$
      mdot v_x + k v_x = 0
      $$



      solving this we obtain



      $$
      v_x = v_x^0 e^-frac km t
      $$



      but according to the initial conditions $v_x^0 = 10$ and also



      $$
      -F_0 = m dot v_x(0) = -m frac km v_x^0Rightarrow 1.2 = k v_x^0 = k times 10
      $$



      so we have



      $$
      k = 0.12Rightarrow frac km = 0.3
      $$



      hence the horizontal dynamics are dictated by



      $$
      dot v_x = -0.3 v_x
      $$






      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        To have a compatible answer with the proposed question we have to consider the movement as horizontal.



        After entering horizontally, the horizontal dynamics are described as



        $$
        m ddot x + kdot x = 0
        $$



        or



        $$
        mdot v_x + k v_x = 0
        $$



        solving this we obtain



        $$
        v_x = v_x^0 e^-frac km t
        $$



        but according to the initial conditions $v_x^0 = 10$ and also



        $$
        -F_0 = m dot v_x(0) = -m frac km v_x^0Rightarrow 1.2 = k v_x^0 = k times 10
        $$



        so we have



        $$
        k = 0.12Rightarrow frac km = 0.3
        $$



        hence the horizontal dynamics are dictated by



        $$
        dot v_x = -0.3 v_x
        $$






        share|cite|improve this answer















        To have a compatible answer with the proposed question we have to consider the movement as horizontal.



        After entering horizontally, the horizontal dynamics are described as



        $$
        m ddot x + kdot x = 0
        $$



        or



        $$
        mdot v_x + k v_x = 0
        $$



        solving this we obtain



        $$
        v_x = v_x^0 e^-frac km t
        $$



        but according to the initial conditions $v_x^0 = 10$ and also



        $$
        -F_0 = m dot v_x(0) = -m frac km v_x^0Rightarrow 1.2 = k v_x^0 = k times 10
        $$



        so we have



        $$
        k = 0.12Rightarrow frac km = 0.3
        $$



        hence the horizontal dynamics are dictated by



        $$
        dot v_x = -0.3 v_x
        $$







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago


























        answered 2 days ago









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