Ok, a body is falling with a velocity $70\sqrt{2}\ ms^{-1}$ and its final velocity must be 0 $ms^{-1}$ after falling a further 200 $m$. What is the acceleration required to bring it to a stop after $200\ m$?

Using the formula $$v^2=v_0^2+2ax$$ and substituting in the known quantities gives $$0=9800+(2a\times200).$$

Rearranging for $a$ gives

\begin{align*} a&=-\frac{9800}{400}\\\\ &= -24.5\ ms^{-2}. \end{align*}

Ok that's easy enough and that's the given answer, but should we not account for $g$ in the answer also?

I mean the acceleration due to gravity hasn't vanished. 'It wants' to speed the body up, so shouldn't the deceleration be $(24.5+g)\ ms^{-2}$ ?

  • $\begingroup$ No, $a$ really is the net deceleration to reduce speed to zero over 200 m of fall. $\endgroup$ – Gert Sep 9 '15 at 15:53

I think you're mentally confusing "acceleration" and "force.

I think your thought process is "If I'm going to apply an acceleration to an object, that acceleration will be fighting against the acceleration of gravity, so the two accelerations will partially cancel each other out".

The problem is that acceleration isn't something you can "apply" to an object - you can apply a force. The acceleration is just something you can measure about an object's motion. So, you're not "applying" an acceleration of $-24.5 ~\rm ms^{-2}$, the object is, according to the problem, accelerating at that rate.

If the problem asked you to find the force necessary to accomplish this, then you would indeed need to account for gravity, because the force that would result in such an acceleration would have to be larger since it's fighting against the influence of gravity:

$$F_{net} = m a$$ $$\Sigma F = F_{applied} + F_g = m a$$ $$F_{applied} = m a - F_g$$ $$F_{applied} = m a - m g$$

The acceleration is merely a measurement of the motion that is described in the problem; there may be many forces on the object including gravity, but we aren't asked to calculate them.

  • $\begingroup$ as much AS I agree with your arguments, gravity is an acceleration rather than a force. That's why all objects, regardless of their mass, fall down with equal acceleration g. $\endgroup$ – bright magus Oct 3 '15 at 7:43

The equation you're using is a kinematic equation. Kinematics, you might recall, is how we mathematically describe motion; it doesn't deal with the causes of that motion. If an object is accelerating at $70 \sqrt{2} \text{ m/s}$ and decelerating at $-24.5 \text{ m/s}^2$ due to a force or a combination of forces, then it will come to rest after traveling 200 m, full stop.1 This would be equally true if this was a car traveling down the highway and slamming on the brakes.

What's confusing you is that there are two forces acting on the object as it decelerates: gravity and a second external agency. This second force must indeed provide enough force to counteract the force of gravity and then some; and if you go through the math, you'll find out that the force required would be $F = m (24.5 \text{ m/s}^2 + g)$, so that the net force is $F = m (24.5 \text{ m/s}^2)$. ($m$ here is the mass of the object.) But as far as the kinematic equations are concerned, the individual forces don't matter; what matters is the net acceleration that they impart.

Forces are usually considered as part of dynamics, the causes of motion, which (if you're just starting out in physics) you may not have covered quite yet.

1 Pun not intended.

  • $\begingroup$ Nice explanation. $\endgroup$ – Brionius Sep 9 '15 at 16:41

Suppose you apply an upwards force $F$ on the falling body. Gravity produces a downwards force $-mg$ so the net force is:

$$ F_{net} = F - mg $$

The net acceleration is then:

$$ a_{net} = \frac{F_{net}}{m} = \frac{F}{m} - g $$

The acceleration you have calculated is the net acceleration $a_{net}$ given by the equation above.


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