0
$\begingroup$

I believe I have some misconceptions regarding this topic. But here goes: I need some help to make sense of an experiment where I sling an object away with a rubber band, with a force of 5 N.

The object is put onto a surface, behind a rubber band. A thread is suspending the rubber band. The thread is then burnt with a match, and the object is released. Upon this, the object have friction with the surface and the air. It flies for 0,5 seconds and reaches a destination at 0,5 meters.

My data is an object weighing 0,05 kg and an external force of 5 N.

Using the formula a = F/m I land on an acceleration of 100 m/s^2.

However, the object does not move further than 0,5 meters, and it travels in 0,5 seconds. Which gives a speed of 1 m/s. And 100 m/s^2 seems outrages.

So how can I include wind resistance and friction into this little experiment, to make more sense?

Or should one simply state that without any other forces acting upon the object, it's acceleration is 100 m/s^2 ?

Thanks in advance!

Improve this question
$\endgroup$
3
  • $\begingroup$ Not entirely clear on your experiment but if you fling the small object away from you with a rubber band and the impulse lasts only 1/100 of a second during which the object is accelerating at 100 m/s^2, then the object will attain a speed of 1 m/s. $\endgroup$
    – user93237
    Commented Feb 28, 2018 at 18:33
  • $\begingroup$ I'm sorry if I've been unclear regarding the experiment. But the impulse lasted 0,5 seconds. I've edited it in the post now. $\endgroup$
    – RMM
    Commented Feb 28, 2018 at 18:35
  • 1
    $\begingroup$ You would really have to show a diagram of your setup. Also, I'm skeptical of the numbers that you stated. If you have a rubber band exerting a 5N force on a small object and accelerating it at 100 m/s^2 for a full 0.5 seconds, then the rubber band has to be in contact with the object and continuously pushing it over a distance of d=(a t^2)/2 = (100 m/s^2)(0.5 s)^2/2 =12.5 meters, which is an unrealistically large distance for any common rubber band. $\endgroup$
    – user93237
    Commented Feb 28, 2018 at 18:48

1 Answer 1

Reset to default
1
$\begingroup$

A large acceleration does not necessarily cause a large velocity: it depends on the time over which that acceleration acts. If your object starts at rest and experiences an acceleration of 100 m/s$^2$ for 0.01 s, its final velocity will be 1 m/s.

However, it sounds as though your object slides with friction after being started, coming to rest in 0.5 s after moving 0.5 m. This means the average velocity is 1 m/s, but assuming a constant frictional force, the instantaneous velocity at the beginning of the slide is 2 m/s. For an acceleration of 100 m/s$^2$, this would imply that the accelerating force was applied for 0.02 s.

Given the method of applying force just added to the question, it is clear that the acceleration will not be constant, but will start at a value of 100 m/s$^2$, and decrease linearly to zero as the rubber band approaches its unstretched length. You can assess the validity of this model by measuring the distance the rubber band is stretched (distance from extended but unstretched point to release point).

Improve this answer
$\endgroup$

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