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If we hit a box (mass = $2$ kg, at rest, without friction) with a force of $2$ Newtons, how do we calculate the acceleration of the box upon hitting using the second law of newton?

I tried this: $f = m*a$

$a = \frac{f}{m}$

=> $a = \frac{2}{2} =1$ m/s^2

but that not right because the acceleration in the beginning must be a bigger number.

Aymen
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  • Why must the acceleration be a "big number"? $F = ma$. $2 = 2 \times ?$ Also, $1$ m/s² = $3.6$ kmph/s. In $100$ s ($1$ minute $40$ seconds), the object would attain a speed of $360$ kmph. That's a "big number", cogito. – Agent Smith Dec 29 '23 at 10:30

1 Answers1

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Your answer is correct. The applied force is so small that acceleration in the beginning will be small too.

Given that,

Mass = 2kg,

Applied force = 2N

We know that, $Acceleration = Applied force/ Mass$

So, acceleration will be = $2/2$ $m/s^2$ = $1$ $m/s^2$

Samin Mustakim
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  • how 1m/s^2 and when we hit the box the velocity changes from 0 to a specific velocity in a very very short time? – Aymen Dec 29 '23 at 10:26
  • Your applied force is too small compared to the mass of the box. So, it isn't possible to have a great acceleration. In 1st second, the velocity will change from 0 to 1$m/s$ – Samin Mustakim Dec 29 '23 at 10:30
  • thank you i understand now, can u tell me how to calculate the initial velocity please at hiting time? i tried: a=f/M; then with integration => v(t) - v(0) = f/m * t; => v(0) = f/m * t +v(t), but we can't here – Aymen Dec 29 '23 at 10:48
  • I saw your post, if I can solve this I will answer that into that post. – Samin Mustakim Dec 29 '23 at 10:51
  • thank you ..... – Aymen Dec 29 '23 at 10:53