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Consider the following diagram:

Reference

It is given that:
A is the position of maximum compression of spring and the child is at rest(at that instant).
At B, the child is having an upward velocity and the spring is relaxed.
At C, the child is again at rest(at that instant).

Now, I have the following doubts:

  1. The author has mentioned that F(net) = 0 at some point below the reference line(not at A). Why is it so? I think that since the child is at rest than F(net) should be equal to zero at position A.
  2. Now, if I am convinced that F(net) = 0 at some point between B and A, then below that point Spring force should be greater than mg, then why is the boy moving downside?
  3. To find maximum Kinetic energy, I can do the following : \begin{align} \frac{dv}{dt}=a=0\end{align} But this doesn't guarantee that it would be maximum, it can also give minimum? And where will be that position(above the reference line or below the reference line)? Why?

Please help me out with this.

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  • $\begingroup$ Sounds like your teacher is incorrect. At point A the net force is zero. $\endgroup$
    – joseph h
    Nov 9, 2022 at 6:12
  • $\begingroup$ And at point B? $\endgroup$ Nov 9, 2022 at 9:06
  • $\begingroup$ Is he moving due to inertia? $\endgroup$ Nov 9, 2022 at 9:06

1 Answer 1

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System - child.
External forces - force due to spring and gravitational attraction due to Earth.

The system has a static equilibrium position when the two external forces are equal in magnitude and opposite in direction (ie no net force is acting on the child) and the child is at rest and that position is somewhere between $\bf A$ and $\bf B$.

When the child is bouncing the child passes through that static equilibrium position possesses maximum kinetic energy as during upward motion there is a net upward force on the child before passing through that point, ie increasing kinetic energy, and a net downward force on the child after passing through that point, ie decreasing kinetic energy.

$\bf A$ is the position of maximum compression of spring and the child is at rest(at that instant).
In this position the upward force due to the spring is greater than the gravitational attraction of the Earth, ie there is a net upward force on the child who will accelerate upwards at the point.

At $\bf B$, the child is having an upward velocity and the spring is relaxed. [and] At $\bf C$, the child is again at rest(at that instant).
There is only the constant gravitational force acting downwards during this phase of the motion resulting in a downward acceleration of the child.

[Upward] Spring force should be greater than $\bf mg$ [downward], then why is the boy moving downside?
Because the child had a downward velocity after passing through the static equilibrium position and was being slowed down by the net upward force until reaching $\bf A$ at which point the child's speed was zero.

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  • $\begingroup$ Will the spring be stretched after point P? Or it will just come to its relaxed position at point P and remains in that position during the phase from A to B? $\endgroup$ Nov 10, 2022 at 1:57
  • $\begingroup$ @ShekharDangi - Since the spring is only attached at one end it can only be compressed and cannot be stretched. $\endgroup$
    – Farcher
    Nov 10, 2022 at 7:13

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