Acceleration is zero, for non-zero net force

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Solution 1

A force is applied to a box on a table(lets ignore friction), and the box moves with some constant velocity.

It's impossible. Or, don't ignore friction.

When an object moves with constant velocity, the total net force on the object is always zero. If you have applied force, there's another force (or, many forces) like friction to counterbalance it.

Another thing I can think of: This argument is missing data. If constant velocity is recorded with respect to table, then there's inertial force to balance your force on box. Meaning, table reference frame is non-inertial.

Solution 2

Maybe you are pouring sand on your box.$$F=\frac{dp}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$$

$$\text{As, } v=0 ms^{-1}$$$$F=v\frac{dm}{dt}$$

Second possibility : If your box is spherical,

By Stokes' Law $$F_{viscous}=6\pi\eta rv$$ where $\eta$ is coefficient of viscosity.

Hence, your ball attains terminal velocity.

$$F=6\pi\eta rv$$

$$v=\frac{F}{6\pi\eta r}$$

Solution 3

By Newton's second law of motion, if there is a nonzero net force there is an acceleration. If there is no acceleration then the net force is zero. In the situation you describe, where the box has no acceleration, there must be another force balancing $F_{app}$ otherwise there will be an acceleration.

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Shinobii
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Updated on November 07, 2020

Comments

  • Shinobii
    Shinobii about 3 years

    I understand a problem like this has already been asked, but I have not found an answer that makes it clear. A force is applied to a box on a table(lets ignore friction), and the box moves with some constant velocity. In this case, acceleration is equal to zero. The net force does not. How can we explain this?

    $$ \sum F = F_{app} = m a $$

    The acceleration $a = 0$, but $F_{app} > 0$.

    Can someone clarify this for me?

    • Carl Witthoft
      Carl Witthoft over 9 years
      Yeah, the question is wrong. Either there is a resisting force such as friction, or the box accelerates. What's your source for this statement/question?
    • Shinobii
      Shinobii over 9 years
      Thought problem. Imagine pushing a box at a constant velocity. Even with friction, $F_{app} > F_{f}$ and $F_{net} > 0$, but the box would still not be accelerating. Would it be safe to assume that IF there is a force acting, there MUST be acceleration ALWAYS?
    • Kenshin
      Kenshin over 9 years
      The answer is simple, the table is on an angle and you are pushing upwards against gravity.
    • user12262
      user12262 over 9 years
      To expand on "Mew's simple solution" (in the above comment) "the table is on an angle and you are pushing upwards against gravity": $\mathbf F_{app} := m \, (\mathbf g\cdot\mathbf v)\mathbf v/\text v^2$; $F_{app} := m \, g \, \text{Sin}[ \phi ]$. If it seems somehow counter-intuitive that the applied force doesn't depend on the velocity $\mathbf v$ remember that the power does: $P = (\mathbf F_{app}\cdot\mathbf v) = F_{app} \, \text v$. (But of course it remains to be seen whether the OP had this sort of "solution" in mind at all.)
    • Vidyanshu Mishra
      Vidyanshu Mishra almost 7 years
      What is the problem if I apply a force just to start motion and then not.(the box will still move with constant velocity since there is no friction)
    • Jagerber48
      Jagerber48 almost 7 years
      @Shinobii Do you mean to say a force is applied briefly and then removed? In that case during the period of time that the force is applied the object is accelerated but after the force is removed the object moves at a constant velocity. The key is that the acceleration at any given moment is proportional to the force at that same given moment.
    • Shinobii
      Shinobii almost 7 years
      I am impressed how much discussion this question stirred up. The question posed was as simple as it sounds. Obviously, the high school textbook that I got this from was crazy. I ended up finding ~40% of the problems were either wrong, or very poorly worded. Do not over think it. Further, to the person who down voted this post, look at the discussion I created. Nothing worse than down voting perfectly good questions. . . People these days.
  • Shinobii
    Shinobii over 9 years
    Yeppers. Rookie mistake. The question was worded incorrectly (cannot rely on the Authors these days). Thanks for your response!
  • Shinobii
    Shinobii over 9 years
    Seems slightly off topic, but interesting reply!
  • evil999man
    evil999man over 9 years
    @Shinobii Note that $F=ma$ is only a special case of $F=dp/dt$