Answer
$25.5\;\rm m/s$
Work Step by Step
We are given the force exerted on the object which is given by the following formula, as the author told us
$$F_x=10\sin\left(\dfrac{2\pi }{4}t\right)\tag 1$$
We know that the particle is pushed by this force from rest. So to find its speed after 2 s of applying this force, we need to use the impulse-momentum theorem.
$$\Delta p_x=J_x$$
where the impulse is given by $J_x=\int_0^tF_xdt$, and the change in momentum is given by $\Delta p_x=mv_{fx}-mv_{ix}=mv_{fx}$ since $v_{ix}=0$.
So,
$$mv_{fx}=\int_0^tF_xdt $$
$$ v_{fx}=\dfrac{1}{m}\int_0^tF_xdt $$
Plugging from (1);
$$ v_{fx}=\dfrac{1}{m}\int_0^t\left[10\sin\left(\dfrac{2\pi }{4}t\right)\right]dt =\dfrac{10}{m}\int_0^t\left[ \sin\left(\dfrac{2\pi }{4}t\right)\right]dt $$
$$ v_{fx}= \dfrac{10}{m}\dfrac{4}{2\pi } \left[ -\cos\left(\dfrac{2\pi }{4}t\right)\right]_0^2
$$
$$v_{fx}= \dfrac{-40}{2m\pi } \left[\overbrace{ \cos\left(\dfrac{2\pi }{4} \times 2\right)}^{ 0}-\overbrace{\cos\left(\dfrac{2\pi }{4} \times 0\right)}^{ 1}\right]=\dfrac{40}{2m\pi}$$
$$v_{fx}=\dfrac{40}{2\times 0.250\;\pi}=\color{red}{\bf 25.5}\;\rm m/s$$
