\(\mathrm{a}=\frac{v-u}{\mathrm{t}}=\left(\frac{40-10}{4}\right) \mathrm{m} / \mathrm{s}^2=7.5 \mathrm{~m} / \mathrm{s}^2\)
one has to be careful here!
\(f eq m a\)
you can't get the required upward force by multiplying the mass and the acceleration. This would have been the case if it were translational (horizontal) motion.
The force to be applied must be sufficient to give the calculated upward acceleration as well as annihilate the downward gravity drag.
Then,
We write Fnet=ma i.e
\begin{aligned}
\mathrm{f}_{\mathrm{u}}-\mathrm{f}_{\mathrm{g}} &=m a \\
\mathrm{f}_{\mathrm{u}} &=m a+\mathrm{f}_{\mathrm{g}} \\
&=m a+\mathrm{m} \\
&=m a+m g \\
&=m(a+g) \\
&=2(7.5+10) \\
&=2(17.5) \\
&=35 \mathrm{~N}
\end{aligned}