Harder, better, faster, stronger

We know, $\dfrac{dF}{dt} = 10 \\ i.e. \displaystyle \int_0^{10} dF = 10 \displaystyle \int_0 ^ t dt\\ i.e. t=1s$ Now, $F=ma$
$\implies a=\dfrac{10}{m}t$
but, $a =\dfrac{dv}{dt}$ So, $\displaystyle \int_0^{50} dv =\displaystyle \int_0^ 1 \dfrac{10t}{m}dt \\ \implies 50 =\dfrac{5}{m} \\ \implies \boxed{m=0.1}$

We know that the force increases linearly in time from 0 to 10 $N$ at a rate of $\gamma=10$ N/s, which takes one second. Then, we have

$\begin{aligned} F &= ma \\ &= m\frac{dv}{dt} \\ &= \gamma t \end{aligned}$

We could use calculus here, or just notice that the relationship between changing velocity and a linearly increasing acceleration is the same as that between changing position and a linearly increasing velocity. Thus, we can peel off the solution $v(t) = \frac{1}{2m}\gamma t^2$

Since the mass accelerates for one second, and its final velocity is 50 m/s, we find $m=\frac{1}{10}$ kg.

Actually you don't have to use calculus for this problem. You can use this method, which is slightly easier if you can see it. we firstly find the time that the force exerts for.

$t=\frac { 10N-0N }{ 10N{ s }^{ -1 } } \\ =1.0s$

Thus if we visualise a graph of Newtons against time(seconds), the area under graph will provide us with the value of Newton seconds.

$[under\quad graph]=\frac { 1 }{ 2 } \times (10N-0N)\times 1.0s\\ \qquad \qquad \qquad \qquad =5.0Ns$

By dividing by the total number of time taken, we can obtain the total force acting on the object.

$\sum F=\frac { 5.0Ns }{ 1.0s } \\ \qquad \qquad =5.0N$

And via the acceleration equation

$a=\frac { \triangle v }{ t } \\ \quad =\frac { 50m{ s }^{ -1 }-0m{ s }^{ -1 } }{ 1.0s } \\ \quad =50m{ s }^{ -2 }$

And as there are no opposing forces,

${ F }_{ net }=5.0N$

And via net force equation

$m=\frac { { F }_{ net } }{ a } \\ \quad =\frac { 5.0N }{ 50m{ s }^{ -2 } } \\ \quad =0.1kg$