Blocks faster than sound

I can't believe that this young person can phrase in such a way. Wanted to report his fault but I tried with one more integer. To my surprise, this was what being meant!

The only graph of acceleration verse time can be drawn which strictly taking his descriptions to be all correct is this:

0 second: 1 m/ s^2;

After

1 second: 1 m/ s^2;

2 second: 8 m/ s^2;

3 second: 15 m/ s^2;

4 second: 22 m/ s^2;

...

58 second: 400 m/ s^2;

59 second: 407 m/ s^2;

60 second: 414 m/ s^2.

All multiplied by 5 kg. A sum of forces is 5 (1 + 8 + 15 + 22 + ... 400 + 407 + 414) = 62250.

Not an average of force over 60 s but a sum of forces! Primitive but usually not asked in such a way. All other ways of thinking just can't take 7 m/ s^3 but 7 m/ s^2, no integer answer or wrong concept to wordings; this includes 1055, 1020+7/ 24 and also 2105 as the answer. Good imagination and I would like to congratulate to Bruno Oggioni Moura. You can think in other ways.

Answer: 62250

The force that was initially applied can be expressed as $F_{0} = m.a \Rightarrow F_{0} = 5 N$ . The sum of all the forces (considering only the seconds, not the seconds' fractions) is $S = \sum_{i = 1}^{j} a_{i}m = m \sum_{i = 1}^{j} a_{i}$ , where $j$ is the total time. So, the sum is $S = 5 \sum_{i = 1}^{60} a_{i} = 5 \times \frac{(a_{1} + a_{60})}{2} \times 60$ .

As $a_{60} = a_{1} + (60 - 1) \tau = 1 +59 \times 7 = 414 m/s^{2}$ , where $\tau$ is the acceleration's variation, the sum of the forces is $S = 5 \times \frac{(1 + 414)}{2} \times 60 = 62250 \frac{kg \times m}{s^{2}} = \boxed{62250 N}$