2
$\begingroup$

I have a really basic question.

If we release an object from a drop tower $700\, \mathrm{m}$ tall, the object free falls for $500\,\mathrm{m}$ and reaches a velocity of $99\mathrm{\frac{m}{s}}$.

For the last $200\mathrm{m}$, a deceleration is applied. The deceleration required to reduce the velocity to zero is calculated to be about $-24 \mathrm{\frac{m}{s^2}}$.

I can't get my head around why we don't need to add the gravitational acceleration $9.81\mathrm{\frac{m}{s^2}}$ on top of this?

I understand we need $-24\mathrm{\frac{m}{s^2}}$ to bring an object traveling at $99\mathrm{\frac{m}{s}}$ to zero velocity over a $200\,\mathrm{m}$ distance, but this is if the object is travelling at a constant speed of $99\mathrm{\frac{m}{s}}$ to start with. In our case, the object is still affected by the gravitational acceleration, don't we need to account for that? $-24 - 9.81 =$ a total deceleration of $-34\mathrm{\frac{m}{s^2}}$?

$\endgroup$
0
$\begingroup$

The first thing to note is that your question is in the realm of kinematics (the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion) so the fact that there is a gravitational force is totally irrelevant when answering the question.

Given the assumption that all the accelerations are constant and down is defined as the positive direction then using the constant acceleration kinematic equations the answer to the question is that the acceleration of the body is $-24 \,\rm m\, s^{-2}$.


If you wish to apply Newton's second law $F=ma$ to the problem then you need to proceed as follows with down as positive noting that the acceleration of the body is still $-24 \,\rm m\, s^{-2}$.

$$F+mg = m\left ( -24 \,\rm m\, s^{-2} \right)$$

where $F$ is the force on the body and $m$ is the mass of the body and all the forces which are acting on the body are on the left hand side of the equation.

With $g = 10 \,\rm m\, s^{-2}$ this results in the force acting on the body equal to $m(-24-10) = -m \,34 \,\rm N$.
This the (upward) force which is required to stop the body in the gravitational field of the Earth.


With no other forces present if a force of this magnitude had been applied to a body of mass $m$ then its acceleration could been found by applying Newton's second law $m \, (-34) = m \,a \Rightarrow a = -34 \,\rm m\, s^{-2}$

$\endgroup$
2
$\begingroup$

$-24.5m/s^2$ is the total acceleration. If you make a force diagram, you will see that in addition to gravitational force, you would have to apply whatever else remaining to make total acceleration be $-24.5m/s^2$. Since gravity is applying acceleration of $9.81m/s^2$, you would need to exert $-34.31m/s^2$ acceleration through external force. Essentially you can make acceleration whatever you want by exerting certain force, and for the ball to reach $0m/s$ at $0m$ the total acceleration should be $-24.5m/s^2$.

(also, I am assuming zero deceleration due to medium friction (such as air)).

$\endgroup$
  • $\begingroup$ Thanks for your reply. I guess both -24.5 and -34.31 are correct. One is the net acceleration, the other is the total acceleration needed to be generated externally. $\endgroup$ – JerryH Oct 23 '18 at 4:08
0
$\begingroup$

We need to look carefully at Newton's equation.

There is a body falling towards the earth with constant acceleration. We could think of it as being produced by another planet, so Newton's equation is:

\begin{align}\require{cancel}F&=m\,a\\\cancel{m}\,a&=\cancel{m}\,{(g-x)}\end{align}

$-x$ mean that the force is opposed to gravity. That's my way to see it as a mass independent trajectory.

So we end with the same equations of motion, where the net acceleration felt by the object is $g-x$. Then as you supposed, this planetary deceleration ($x$) is $a - g$, i.e $-24-10\approx -34$

$\endgroup$
  • $\begingroup$ Thanks for the reply. This is what I thought. We will need to generate -34 m/s^2. $\endgroup$ – JerryH Oct 23 '18 at 4:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.