Question

What is the value of work done on an object when a 70–newton force moves it
9.0 meters in the same direction as the force?

Answer 1

Work, very simply, equals force times distance (when the force and distance are in the same direction. otherwise you get a little bit of trig added on)

W=F∗Δx

W=70N * 9.0 m = 630 Nm = 630 J

Answer 2

You should also know that the true equation for work is the dot product of force and distance.
\[\huge W=\vec{F} \cdot \vec{d}\]
If you're familiar with dot product, then we can further express this as
\[\huge W=|\vec{F}||\vec{d}| \cos(\theta)\]
Where \(\theta\) is the angle between the force and position vectors.

In your scenario, you're already given the magnitudes (since we're not given in vector form). So we can drop the magnitude symbols. Additionally, we're given that the force applied is in the SAME direction as the direction that the box is moving! Therefore, \(\theta=0\) and \(\cos(0)=1\). Your final simplification for your equation is now:
\[\huge W=Fd\]
which is the result that @kitties_rules got